Mixing
show this inequality with $sum_{i=1}^{n}a_{i}=n$
$begingroup$
Let $nge 3$ be postive number,$a_{i}>0,i=1,2,cdots,n$,and $displaystylesum_{i=1}^{n}a_{i}=n$,show that
$$a^3_{1}a_{2}+a^3_{2}a_{3}+cdots+a^3_{n}a_{1}+nge 2(a_{1}a_{2}cdots a_{n-1}+a_{2}a_{3}cdots a_{n}+a_{n}a_{1}cdots a_{n-2})$$
it seem can use indution to prove it.when $n=3$,it must prove
$$a^3_{1}a_{2}+a^3_{2}a_{3}+a^3_{3}a_{1}+3ge 2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
it seem use three shcur inequaliy
$$a^3+b^3+c^3+3abcge sum ab(a+b)$$
then we have
$$a^2+b^2+c^2+3(abc)^{2/3}ge 2(ab+bc+ca)$$
multivariable-calculus inequality a.m.-g.m.-inequality buffalo-way
edited Jan 25 at 17:58
Michael Rozenberg
101k1591193
asked Jan 10 at 3:35
geromtygeromty
977423
$endgroup$
add a comment |
$begingroup$
Let $nge 3$ be postive number,$a_{i}>0,i=1,2,cdots,n$,and $displaystylesum_{i=1}^{n}a_{i}=n$,show that
$$a^3_{1}a_{2}+a^3_{2}a_{3}+cdots+a^3_{n}a_{1}+nge 2(a_{1}a_{2}cdots a_{n-1}+a_{2}a_{3}cdots a_{n}+a_{n}a_{1}cdots a_{n-2})$$
it seem can use indution to prove it.when $n=3$,it must prove
$$a^3_{1}a_{2}+a^3_{2}a_{3}+a^3_{3}a_{1}+3ge 2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
it seem use three shcur inequaliy
$$a^3+b^3+c^3+3abcge sum ab(a+b)$$
then we have
$$a^2+b^2+c^2+3(abc)^{2/3}ge 2(ab+bc+ca)$$
multivariable-calculus inequality a.m.-g.m.-inequality buffalo-way
edited Jan 25 at 17:58
Michael Rozenberg
101k1591193
asked Jan 10 at 3:35
geromtygeromty
977423
$endgroup$
$begingroup$
Do you have a source for this problem? Where did you find it? Please mention for context.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 4:00
$begingroup$
For $n=4$ how is the inequality formed?
$endgroup$
– Nikos Bagis
Jan 20 at 23:13
$begingroup$
@Nikos Bagis Let $a$, $b$, $c$ and $d$ be positive numbers such that $a+b+c+d=4$. Prove that: $a^3b+b^3c+c^3d+d^3a+4geq2(abc+abd+acd+bcd).$ It's true, I proved it.
$endgroup$
– Michael Rozenberg
Jan 21 at 3:41
$begingroup$
So there is misprint in the statement in the given inequality. It should read as $a_1^3a_2+a_2^3a_3+ldots+a_n^3a_1+ngeq 2sum^{n}_{i_1<i_2<ldots<i_{n-1}}a_{i_1}a_{i_2}ldots a_{i_{n-1}}$
$endgroup$
– Nikos Bagis
Jan 21 at 8:31
add a comment |
$begingroup$
Let $nge 3$ be postive number,$a_{i}>0,i=1,2,cdots,n$,and $displaystylesum_{i=1}^{n}a_{i}=n$,show that
$$a^3_{1}a_{2}+a^3_{2}a_{3}+cdots+a^3_{n}a_{1}+nge 2(a_{1}a_{2}cdots a_{n-1}+a_{2}a_{3}cdots a_{n}+a_{n}a_{1}cdots a_{n-2})$$
it seem can use indution to prove it.when $n=3$,it must prove
$$a^3_{1}a_{2}+a^3_{2}a_{3}+a^3_{3}a_{1}+3ge 2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
it seem use three shcur inequaliy
$$a^3+b^3+c^3+3abcge sum ab(a+b)$$
then we have
$$a^2+b^2+c^2+3(abc)^{2/3}ge 2(ab+bc+ca)$$
multivariable-calculus inequality a.m.-g.m.-inequality buffalo-way
edited Jan 25 at 17:58
Michael Rozenberg
101k1591193
asked Jan 10 at 3:35
geromtygeromty
977423
$endgroup$
Let $nge 3$ be postive number,$a_{i}>0,i=1,2,cdots,n$,and $displaystylesum_{i=1}^{n}a_{i}=n$,show that
$$a^3_{1}a_{2}+a^3_{2}a_{3}+cdots+a^3_{n}a_{1}+nge 2(a_{1}a_{2}cdots a_{n-1}+a_{2}a_{3}cdots a_{n}+a_{n}a_{1}cdots a_{n-2})$$
it seem can use indution to prove it.when $n=3$,it must prove
$$a^3_{1}a_{2}+a^3_{2}a_{3}+a^3_{3}a_{1}+3ge 2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})$$
it seem use three shcur inequaliy
$$a^3+b^3+c^3+3abcge sum ab(a+b)$$
then we have
$$a^2+b^2+c^2+3(abc)^{2/3}ge 2(ab+bc+ca)$$
multivariable-calculus inequality a.m.-g.m.-inequality buffalo-way
multivariable-calculus inequality a.m.-g.m.-inequality buffalo-way
edited Jan 25 at 17:58
Michael Rozenberg
101k1591193
asked Jan 10 at 3:35
geromtygeromty
977423
edited Jan 25 at 17:58
Michael Rozenberg
101k1591193
asked Jan 10 at 3:35
geromtygeromty
977423
edited Jan 25 at 17:58
Michael Rozenberg
101k1591193
edited Jan 25 at 17:58
Michael Rozenberg
101k1591193
edited Jan 25 at 17:58
Michael Rozenberg
101k1591193
101k1591193
asked Jan 10 at 3:35
geromtygeromty
977423
asked Jan 10 at 3:35
geromtygeromty
977423
asked Jan 10 at 3:35
geromtygeromty
977423
977423
$begingroup$
Do you have a source for this problem? Where did you find it? Please mention for context.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 4:00
$begingroup$
For $n=4$ how is the inequality formed?
$endgroup$
– Nikos Bagis
Jan 20 at 23:13
$begingroup$
@Nikos Bagis Let $a$, $b$, $c$ and $d$ be positive numbers such that $a+b+c+d=4$. Prove that: $a^3b+b^3c+c^3d+d^3a+4geq2(abc+abd+acd+bcd).$ It's true, I proved it.
$endgroup$
– Michael Rozenberg
Jan 21 at 3:41
$begingroup$
So there is misprint in the statement in the given inequality. It should read as $a_1^3a_2+a_2^3a_3+ldots+a_n^3a_1+ngeq 2sum^{n}_{i_1<i_2<ldots<i_{n-1}}a_{i_1}a_{i_2}ldots a_{i_{n-1}}$
$endgroup$
– Nikos Bagis
Jan 21 at 8:31
add a comment |
$begingroup$
Do you have a source for this problem? Where did you find it? Please mention for context.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 4:00
$begingroup$
For $n=4$ how is the inequality formed?
$endgroup$
– Nikos Bagis
Jan 20 at 23:13
$begingroup$
@Nikos Bagis Let $a$, $b$, $c$ and $d$ be positive numbers such that $a+b+c+d=4$. Prove that: $a^3b+b^3c+c^3d+d^3a+4geq2(abc+abd+acd+bcd).$ It's true, I proved it.
$endgroup$
– Michael Rozenberg
Jan 21 at 3:41
$begingroup$
So there is misprint in the statement in the given inequality. It should read as $a_1^3a_2+a_2^3a_3+ldots+a_n^3a_1+ngeq 2sum^{n}_{i_1<i_2<ldots<i_{n-1}}a_{i_1}a_{i_2}ldots a_{i_{n-1}}$
$endgroup$
– Nikos Bagis
Jan 21 at 8:31
$begingroup$
Do you have a source for this problem? Where did you find it? Please mention for context.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 4:00
$begingroup$
Do you have a source for this problem? Where did you find it? Please mention for context.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 4:00
$begingroup$
For $n=4$ how is the inequality formed?
$endgroup$
– Nikos Bagis
Jan 20 at 23:13
$begingroup$
For $n=4$ how is the inequality formed?
$endgroup$
– Nikos Bagis
Jan 20 at 23:13
$begingroup$
@Nikos Bagis Let $a$, $b$, $c$ and $d$ be positive numbers such that $a+b+c+d=4$. Prove that: $a^3b+b^3c+c^3d+d^3a+4geq2(abc+abd+acd+bcd).$ It's true, I proved it.
$endgroup$
– Michael Rozenberg
Jan 21 at 3:41
$begingroup$
@Nikos Bagis Let $a$, $b$, $c$ and $d$ be positive numbers such that $a+b+c+d=4$. Prove that: $a^3b+b^3c+c^3d+d^3a+4geq2(abc+abd+acd+bcd).$ It's true, I proved it.
$endgroup$
– Michael Rozenberg
Jan 21 at 3:41
$begingroup$
So there is misprint in the statement in the given inequality. It should read as $a_1^3a_2+a_2^3a_3+ldots+a_n^3a_1+ngeq 2sum^{n}_{i_1<i_2<ldots<i_{n-1}}a_{i_1}a_{i_2}ldots a_{i_{n-1}}$
$endgroup$
– Nikos Bagis
Jan 21 at 8:31
$begingroup$
So there is misprint in the statement in the given inequality. It should read as $a_1^3a_2+a_2^3a_3+ldots+a_n^3a_1+ngeq 2sum^{n}_{i_1<i_2<ldots<i_{n-1}}a_{i_1}a_{i_2}ldots a_{i_{n-1}}$
$endgroup$
– Nikos Bagis
Jan 21 at 8:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The hint.
Prove this inequality for $n=3$ and for $n=4$.
But for all $ngeq5$ by AM-GM we obtain:
$$sum_{k=1}^na_k^3a_{k+1}geq nsqrt[n]{prod_{k=1}^na_k^4}.$$
Thus, it's enough to prove that:
$$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
Now, let $prodlimits_{k=1}^na_k=const$ and $f(x)=-frac{1}{x}.$
Thus, $$g(x)=f'left(frac{1}{x}right)=x^2$$ is strictly convex on $(0,+infty),$ which says that by Vasc's EV Method, theorem 1.3 (1), $p=0$ from here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
the expression $sumlimits_{k=1}^nf(x_k)=-sumlimits_{k=1}^nfrac{1}{a_k}$ gets a minimal value, when $n-1$ variables are equal.
After homogenization it's enough to assume that $a_1=a$ and $a_2=...=a_n=1$ and we need to prove that:
$$nsqrt[n]{a^4}left(frac{a+n-1}{n}right)^{n-5}+nleft(frac{a+n-1}{n}right)^{n-1}geq2((n-1)a+1),$$
which is true by AM-GM!
Indeed, let $a=x^n$.
Thus, we need to prove that
$$P(x)=x^{n^2-n}+b_1x^{n^2-n-1}+...+b_{n^2-n}geq0,$$ where only coefficient before $x^n$ is negative.
But the inequality $$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
is symmetric, which says that the polynomial $P$ is divisible by $(x-1)^2.$
Now, we see that the sum of coefficients of $P'(x)$ is equal to zero,
which says that $P(x)geq0$ by AM-GM.
answered Jan 21 at 20:37
Michael RozenbergMichael Rozenberg
101k1591193
$endgroup$
$begingroup$
How is your last inequality true?
$endgroup$
– didgogns
Jan 21 at 22:32
$begingroup$
@didgogns I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 22 at 8:55
1
$begingroup$
A great idea, I struggled with second derivatives and $(1+x)^nge1+xn)$ all day but this method is better!
$endgroup$
– didgogns
Jan 22 at 9:07
$begingroup$
but how to prove $n=3$ and $n=4$?
$endgroup$
– geromty
Jan 23 at 10:31
$begingroup$
@geromty I proved it by BW. For $n=3$ we can use also the $uvw$'s technique, but BW is much more better here.
$endgroup$
– Michael Rozenberg
Jan 23 at 10:35
add a comment |
$begingroup$
A dull proof for $n<5$ by BW(Buffalo Way).
For $n=4$, it is enough to prove$$64(a^3b+b^3c+c^3d+d^3a)+(a+b+c+d)^2 ge 32(a+b+c+d)(abc+abd+acd+bcd)$$and WLOG we can assume $a$ is smallest of the four numbers.
That is, $b=a+u$, $c=a+v$ and $d=a+w$ for some positive numbers $u,v,w$. Expanding gives
$$64a^2(3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2)\
+16aleft(left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w right)\
+left(left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w right)ge 0$$
and each coeffcients are nonnegative. Indeed,
$$3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2 ge 2(u-w)^2+0.5(u-v)^2+0.5(v-w)^2 ge 0,$$
$$left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w\ge left(sum_{cyc} 4u^3right) + 4 u^2 v - u w^2 -7vw^2 + 11 v^2 w \ge (4u^3-uw^2+2w^3)+(4v^3+11v^2w-7vw^2+2w^3)ge 0$$
and
$$left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w\geleft(sum_{cyc} 4u^3w-13u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3w\ge v(68u^3-13u^2w-13uw^2+4w^3)+w(68v^3-13uv^2+4u^3)ge 0$$
For $n=3$, it is enough to show
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4ge6(ab+bc+ca)(a+b+c)^2$$
and similarly, let $b=a+u$, $c=a+v$ and expanding gives
$$45a^2(u^2 - u v + v^2)\
+9a(3 u^3 + 5 u^2 v - 4 u v^2 + 3 v^3)\
+(u^4 + 25 u^3 v - 6 u^2 v^2 - 2 u v^3 + v^4)ge 0$$
and you can check each of these polynomials are nonnegative for positive $u, v, w$.
answered Jan 25 at 17:03
didgognsdidgogns
3,218523
$endgroup$
$begingroup$
Yes! The hardest case $ngeq5$ turned out simple.
$endgroup$
– Michael Rozenberg
Jan 25 at 18:00
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The hint.
Prove this inequality for $n=3$ and for $n=4$.
But for all $ngeq5$ by AM-GM we obtain:
$$sum_{k=1}^na_k^3a_{k+1}geq nsqrt[n]{prod_{k=1}^na_k^4}.$$
Thus, it's enough to prove that:
$$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
Now, let $prodlimits_{k=1}^na_k=const$ and $f(x)=-frac{1}{x}.$
Thus, $$g(x)=f'left(frac{1}{x}right)=x^2$$ is strictly convex on $(0,+infty),$ which says that by Vasc's EV Method, theorem 1.3 (1), $p=0$ from here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
the expression $sumlimits_{k=1}^nf(x_k)=-sumlimits_{k=1}^nfrac{1}{a_k}$ gets a minimal value, when $n-1$ variables are equal.
After homogenization it's enough to assume that $a_1=a$ and $a_2=...=a_n=1$ and we need to prove that:
$$nsqrt[n]{a^4}left(frac{a+n-1}{n}right)^{n-5}+nleft(frac{a+n-1}{n}right)^{n-1}geq2((n-1)a+1),$$
which is true by AM-GM!
Indeed, let $a=x^n$.
Thus, we need to prove that
$$P(x)=x^{n^2-n}+b_1x^{n^2-n-1}+...+b_{n^2-n}geq0,$$ where only coefficient before $x^n$ is negative.
But the inequality $$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
is symmetric, which says that the polynomial $P$ is divisible by $(x-1)^2.$
Now, we see that the sum of coefficients of $P'(x)$ is equal to zero,
which says that $P(x)geq0$ by AM-GM.
answered Jan 21 at 20:37
Michael RozenbergMichael Rozenberg
101k1591193
$endgroup$
$begingroup$
How is your last inequality true?
$endgroup$
– didgogns
Jan 21 at 22:32
$begingroup$
@didgogns I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 22 at 8:55
1
$begingroup$
A great idea, I struggled with second derivatives and $(1+x)^nge1+xn)$ all day but this method is better!
$endgroup$
– didgogns
Jan 22 at 9:07
$begingroup$
but how to prove $n=3$ and $n=4$?
$endgroup$
– geromty
Jan 23 at 10:31
$begingroup$
@geromty I proved it by BW. For $n=3$ we can use also the $uvw$'s technique, but BW is much more better here.
$endgroup$
– Michael Rozenberg
Jan 23 at 10:35
add a comment |
$begingroup$
The hint.
Prove this inequality for $n=3$ and for $n=4$.
But for all $ngeq5$ by AM-GM we obtain:
$$sum_{k=1}^na_k^3a_{k+1}geq nsqrt[n]{prod_{k=1}^na_k^4}.$$
Thus, it's enough to prove that:
$$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
Now, let $prodlimits_{k=1}^na_k=const$ and $f(x)=-frac{1}{x}.$
Thus, $$g(x)=f'left(frac{1}{x}right)=x^2$$ is strictly convex on $(0,+infty),$ which says that by Vasc's EV Method, theorem 1.3 (1), $p=0$ from here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
the expression $sumlimits_{k=1}^nf(x_k)=-sumlimits_{k=1}^nfrac{1}{a_k}$ gets a minimal value, when $n-1$ variables are equal.
After homogenization it's enough to assume that $a_1=a$ and $a_2=...=a_n=1$ and we need to prove that:
$$nsqrt[n]{a^4}left(frac{a+n-1}{n}right)^{n-5}+nleft(frac{a+n-1}{n}right)^{n-1}geq2((n-1)a+1),$$
which is true by AM-GM!
Indeed, let $a=x^n$.
Thus, we need to prove that
$$P(x)=x^{n^2-n}+b_1x^{n^2-n-1}+...+b_{n^2-n}geq0,$$ where only coefficient before $x^n$ is negative.
But the inequality $$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
is symmetric, which says that the polynomial $P$ is divisible by $(x-1)^2.$
Now, we see that the sum of coefficients of $P'(x)$ is equal to zero,
which says that $P(x)geq0$ by AM-GM.
answered Jan 21 at 20:37
Michael RozenbergMichael Rozenberg
101k1591193
$endgroup$
$begingroup$
How is your last inequality true?
$endgroup$
– didgogns
Jan 21 at 22:32
$begingroup$
@didgogns I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 22 at 8:55
1
$begingroup$
A great idea, I struggled with second derivatives and $(1+x)^nge1+xn)$ all day but this method is better!
$endgroup$
– didgogns
Jan 22 at 9:07
$begingroup$
but how to prove $n=3$ and $n=4$?
$endgroup$
– geromty
Jan 23 at 10:31
$begingroup$
@geromty I proved it by BW. For $n=3$ we can use also the $uvw$'s technique, but BW is much more better here.
$endgroup$
– Michael Rozenberg
Jan 23 at 10:35
add a comment |
$begingroup$
The hint.
Prove this inequality for $n=3$ and for $n=4$.
But for all $ngeq5$ by AM-GM we obtain:
$$sum_{k=1}^na_k^3a_{k+1}geq nsqrt[n]{prod_{k=1}^na_k^4}.$$
Thus, it's enough to prove that:
$$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
Now, let $prodlimits_{k=1}^na_k=const$ and $f(x)=-frac{1}{x}.$
Thus, $$g(x)=f'left(frac{1}{x}right)=x^2$$ is strictly convex on $(0,+infty),$ which says that by Vasc's EV Method, theorem 1.3 (1), $p=0$ from here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
the expression $sumlimits_{k=1}^nf(x_k)=-sumlimits_{k=1}^nfrac{1}{a_k}$ gets a minimal value, when $n-1$ variables are equal.
After homogenization it's enough to assume that $a_1=a$ and $a_2=...=a_n=1$ and we need to prove that:
$$nsqrt[n]{a^4}left(frac{a+n-1}{n}right)^{n-5}+nleft(frac{a+n-1}{n}right)^{n-1}geq2((n-1)a+1),$$
which is true by AM-GM!
Indeed, let $a=x^n$.
Thus, we need to prove that
$$P(x)=x^{n^2-n}+b_1x^{n^2-n-1}+...+b_{n^2-n}geq0,$$ where only coefficient before $x^n$ is negative.
But the inequality $$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
is symmetric, which says that the polynomial $P$ is divisible by $(x-1)^2.$
Now, we see that the sum of coefficients of $P'(x)$ is equal to zero,
which says that $P(x)geq0$ by AM-GM.
answered Jan 21 at 20:37
Michael RozenbergMichael Rozenberg
101k1591193
$endgroup$
The hint.
Prove this inequality for $n=3$ and for $n=4$.
But for all $ngeq5$ by AM-GM we obtain:
$$sum_{k=1}^na_k^3a_{k+1}geq nsqrt[n]{prod_{k=1}^na_k^4}.$$
Thus, it's enough to prove that:
$$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
Now, let $prodlimits_{k=1}^na_k=const$ and $f(x)=-frac{1}{x}.$
Thus, $$g(x)=f'left(frac{1}{x}right)=x^2$$ is strictly convex on $(0,+infty),$ which says that by Vasc's EV Method, theorem 1.3 (1), $p=0$ from here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
the expression $sumlimits_{k=1}^nf(x_k)=-sumlimits_{k=1}^nfrac{1}{a_k}$ gets a minimal value, when $n-1$ variables are equal.
After homogenization it's enough to assume that $a_1=a$ and $a_2=...=a_n=1$ and we need to prove that:
$$nsqrt[n]{a^4}left(frac{a+n-1}{n}right)^{n-5}+nleft(frac{a+n-1}{n}right)^{n-1}geq2((n-1)a+1),$$
which is true by AM-GM!
Indeed, let $a=x^n$.
Thus, we need to prove that
$$P(x)=x^{n^2-n}+b_1x^{n^2-n-1}+...+b_{n^2-n}geq0,$$ where only coefficient before $x^n$ is negative.
But the inequality $$nsqrt[n]{prod_{k=1}^na_k^4}+ngeq2prod_{k=1}^na_ksum_{k=1}^nfrac{1}{a_k}.$$
is symmetric, which says that the polynomial $P$ is divisible by $(x-1)^2.$
Now, we see that the sum of coefficients of $P'(x)$ is equal to zero,
which says that $P(x)geq0$ by AM-GM.
answered Jan 21 at 20:37
Michael RozenbergMichael Rozenberg
101k1591193
edited Jan 22 at 8:55
answered Jan 21 at 20:37
Michael RozenbergMichael Rozenberg
101k1591193
answered Jan 21 at 20:37
Michael RozenbergMichael Rozenberg
101k1591193
answered Jan 21 at 20:37
Michael RozenbergMichael Rozenberg
101k1591193
101k1591193
$begingroup$
How is your last inequality true?
$endgroup$
– didgogns
Jan 21 at 22:32
$begingroup$
@didgogns I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 22 at 8:55
1
$begingroup$
A great idea, I struggled with second derivatives and $(1+x)^nge1+xn)$ all day but this method is better!
$endgroup$
– didgogns
Jan 22 at 9:07
$begingroup$
but how to prove $n=3$ and $n=4$?
$endgroup$
– geromty
Jan 23 at 10:31
$begingroup$
@geromty I proved it by BW. For $n=3$ we can use also the $uvw$'s technique, but BW is much more better here.
$endgroup$
– Michael Rozenberg
Jan 23 at 10:35
add a comment |
$begingroup$
How is your last inequality true?
$endgroup$
– didgogns
Jan 21 at 22:32
$begingroup$
@didgogns I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 22 at 8:55
1
$begingroup$
A great idea, I struggled with second derivatives and $(1+x)^nge1+xn)$ all day but this method is better!
$endgroup$
– didgogns
Jan 22 at 9:07
$begingroup$
but how to prove $n=3$ and $n=4$?
$endgroup$
– geromty
Jan 23 at 10:31
$begingroup$
@geromty I proved it by BW. For $n=3$ we can use also the $uvw$'s technique, but BW is much more better here.
$endgroup$
– Michael Rozenberg
Jan 23 at 10:35
$begingroup$
How is your last inequality true?
$endgroup$
– didgogns
Jan 21 at 22:32
$begingroup$
How is your last inequality true?
$endgroup$
– didgogns
Jan 21 at 22:32
$begingroup$
@didgogns I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 22 at 8:55
$begingroup$
@didgogns I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 22 at 8:55
1
1
$begingroup$
A great idea, I struggled with second derivatives and $(1+x)^nge1+xn)$ all day but this method is better!
$endgroup$
– didgogns
Jan 22 at 9:07
$begingroup$
A great idea, I struggled with second derivatives and $(1+x)^nge1+xn)$ all day but this method is better!
$endgroup$
– didgogns
Jan 22 at 9:07
$begingroup$
but how to prove $n=3$ and $n=4$?
$endgroup$
– geromty
Jan 23 at 10:31
$begingroup$
but how to prove $n=3$ and $n=4$?
$endgroup$
– geromty
Jan 23 at 10:31
$begingroup$
@geromty I proved it by BW. For $n=3$ we can use also the $uvw$'s technique, but BW is much more better here.
$endgroup$
– Michael Rozenberg
Jan 23 at 10:35
$begingroup$
@geromty I proved it by BW. For $n=3$ we can use also the $uvw$'s technique, but BW is much more better here.
$endgroup$
– Michael Rozenberg
Jan 23 at 10:35
add a comment |
$begingroup$
A dull proof for $n<5$ by BW(Buffalo Way).
For $n=4$, it is enough to prove$$64(a^3b+b^3c+c^3d+d^3a)+(a+b+c+d)^2 ge 32(a+b+c+d)(abc+abd+acd+bcd)$$and WLOG we can assume $a$ is smallest of the four numbers.
That is, $b=a+u$, $c=a+v$ and $d=a+w$ for some positive numbers $u,v,w$. Expanding gives
$$64a^2(3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2)\
+16aleft(left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w right)\
+left(left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w right)ge 0$$
and each coeffcients are nonnegative. Indeed,
$$3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2 ge 2(u-w)^2+0.5(u-v)^2+0.5(v-w)^2 ge 0,$$
$$left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w\ge left(sum_{cyc} 4u^3right) + 4 u^2 v - u w^2 -7vw^2 + 11 v^2 w \ge (4u^3-uw^2+2w^3)+(4v^3+11v^2w-7vw^2+2w^3)ge 0$$
and
$$left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w\geleft(sum_{cyc} 4u^3w-13u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3w\ge v(68u^3-13u^2w-13uw^2+4w^3)+w(68v^3-13uv^2+4u^3)ge 0$$
For $n=3$, it is enough to show
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4ge6(ab+bc+ca)(a+b+c)^2$$
and similarly, let $b=a+u$, $c=a+v$ and expanding gives
$$45a^2(u^2 - u v + v^2)\
+9a(3 u^3 + 5 u^2 v - 4 u v^2 + 3 v^3)\
+(u^4 + 25 u^3 v - 6 u^2 v^2 - 2 u v^3 + v^4)ge 0$$
and you can check each of these polynomials are nonnegative for positive $u, v, w$.
answered Jan 25 at 17:03
didgognsdidgogns
3,218523
$endgroup$
$begingroup$
Yes! The hardest case $ngeq5$ turned out simple.
$endgroup$
– Michael Rozenberg
Jan 25 at 18:00
add a comment |
$begingroup$
A dull proof for $n<5$ by BW(Buffalo Way).
For $n=4$, it is enough to prove$$64(a^3b+b^3c+c^3d+d^3a)+(a+b+c+d)^2 ge 32(a+b+c+d)(abc+abd+acd+bcd)$$and WLOG we can assume $a$ is smallest of the four numbers.
That is, $b=a+u$, $c=a+v$ and $d=a+w$ for some positive numbers $u,v,w$. Expanding gives
$$64a^2(3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2)\
+16aleft(left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w right)\
+left(left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w right)ge 0$$
and each coeffcients are nonnegative. Indeed,
$$3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2 ge 2(u-w)^2+0.5(u-v)^2+0.5(v-w)^2 ge 0,$$
$$left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w\ge left(sum_{cyc} 4u^3right) + 4 u^2 v - u w^2 -7vw^2 + 11 v^2 w \ge (4u^3-uw^2+2w^3)+(4v^3+11v^2w-7vw^2+2w^3)ge 0$$
and
$$left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w\geleft(sum_{cyc} 4u^3w-13u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3w\ge v(68u^3-13u^2w-13uw^2+4w^3)+w(68v^3-13uv^2+4u^3)ge 0$$
For $n=3$, it is enough to show
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4ge6(ab+bc+ca)(a+b+c)^2$$
and similarly, let $b=a+u$, $c=a+v$ and expanding gives
$$45a^2(u^2 - u v + v^2)\
+9a(3 u^3 + 5 u^2 v - 4 u v^2 + 3 v^3)\
+(u^4 + 25 u^3 v - 6 u^2 v^2 - 2 u v^3 + v^4)ge 0$$
and you can check each of these polynomials are nonnegative for positive $u, v, w$.
answered Jan 25 at 17:03
didgognsdidgogns
3,218523
$endgroup$
$begingroup$
Yes! The hardest case $ngeq5$ turned out simple.
$endgroup$
– Michael Rozenberg
Jan 25 at 18:00
add a comment |
$begingroup$
A dull proof for $n<5$ by BW(Buffalo Way).
For $n=4$, it is enough to prove$$64(a^3b+b^3c+c^3d+d^3a)+(a+b+c+d)^2 ge 32(a+b+c+d)(abc+abd+acd+bcd)$$and WLOG we can assume $a$ is smallest of the four numbers.
That is, $b=a+u$, $c=a+v$ and $d=a+w$ for some positive numbers $u,v,w$. Expanding gives
$$64a^2(3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2)\
+16aleft(left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w right)\
+left(left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w right)ge 0$$
and each coeffcients are nonnegative. Indeed,
$$3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2 ge 2(u-w)^2+0.5(u-v)^2+0.5(v-w)^2 ge 0,$$
$$left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w\ge left(sum_{cyc} 4u^3right) + 4 u^2 v - u w^2 -7vw^2 + 11 v^2 w \ge (4u^3-uw^2+2w^3)+(4v^3+11v^2w-7vw^2+2w^3)ge 0$$
and
$$left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w\geleft(sum_{cyc} 4u^3w-13u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3w\ge v(68u^3-13u^2w-13uw^2+4w^3)+w(68v^3-13uv^2+4u^3)ge 0$$
For $n=3$, it is enough to show
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4ge6(ab+bc+ca)(a+b+c)^2$$
and similarly, let $b=a+u$, $c=a+v$ and expanding gives
$$45a^2(u^2 - u v + v^2)\
+9a(3 u^3 + 5 u^2 v - 4 u v^2 + 3 v^3)\
+(u^4 + 25 u^3 v - 6 u^2 v^2 - 2 u v^3 + v^4)ge 0$$
and you can check each of these polynomials are nonnegative for positive $u, v, w$.
answered Jan 25 at 17:03
didgognsdidgogns
3,218523
$endgroup$
A dull proof for $n<5$ by BW(Buffalo Way).
For $n=4$, it is enough to prove$$64(a^3b+b^3c+c^3d+d^3a)+(a+b+c+d)^2 ge 32(a+b+c+d)(abc+abd+acd+bcd)$$and WLOG we can assume $a$ is smallest of the four numbers.
That is, $b=a+u$, $c=a+v$ and $d=a+w$ for some positive numbers $u,v,w$. Expanding gives
$$64a^2(3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2)\
+16aleft(left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w right)\
+left(left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w right)ge 0$$
and each coeffcients are nonnegative. Indeed,
$$3 u^2 - u v - 4 u w + 3v^2 - v w + 3 w^2 ge 2(u-w)^2+0.5(u-v)^2+0.5(v-w)^2 ge 0,$$
$$left(sum_{cyc} 5u^3-u^2wright) + 11 u^2 v - 14 u v w - u w^2 + 11 v^2 w\ge left(sum_{cyc} 4u^3right) + 4 u^2 v - u w^2 -7vw^2 + 11 v^2 w \ge (4u^3-uw^2+2w^3)+(4v^3+11v^2w-7vw^2+2w^3)ge 0$$
and
$$left(sum_{cyc} u^4+6u^2v^2+4u^3w-20u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3 w\geleft(sum_{cyc} 4u^3w-13u^2vwright) + 68 u^3 v + 4 u w^3 + 68 v^3w\ge v(68u^3-13u^2w-13uw^2+4w^3)+w(68v^3-13uv^2+4u^3)ge 0$$
For $n=3$, it is enough to show
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4ge6(ab+bc+ca)(a+b+c)^2$$
and similarly, let $b=a+u$, $c=a+v$ and expanding gives
$$45a^2(u^2 - u v + v^2)\
+9a(3 u^3 + 5 u^2 v - 4 u v^2 + 3 v^3)\
+(u^4 + 25 u^3 v - 6 u^2 v^2 - 2 u v^3 + v^4)ge 0$$
and you can check each of these polynomials are nonnegative for positive $u, v, w$.
answered Jan 25 at 17:03
didgognsdidgogns
3,218523
answered Jan 25 at 17:03
didgognsdidgogns
3,218523
answered Jan 25 at 17:03
didgognsdidgogns
3,218523
answered Jan 25 at 17:03
didgognsdidgogns
3,218523
3,218523
$begingroup$
Yes! The hardest case $ngeq5$ turned out simple.
$endgroup$
– Michael Rozenberg
Jan 25 at 18:00
add a comment |
$begingroup$
Yes! The hardest case $ngeq5$ turned out simple.
$endgroup$
– Michael Rozenberg
Jan 25 at 18:00
$begingroup$
Yes! The hardest case $ngeq5$ turned out simple.
$endgroup$
– Michael Rozenberg
Jan 25 at 18:00
$begingroup$
Yes! The hardest case $ngeq5$ turned out simple.
$endgroup$
– Michael Rozenberg
Jan 25 at 18:00
add a comment |
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$begingroup$
Do you have a source for this problem? Where did you find it? Please mention for context.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 4:00
$begingroup$
For $n=4$ how is the inequality formed?
$endgroup$
– Nikos Bagis
Jan 20 at 23:13
$begingroup$
@Nikos Bagis Let $a$, $b$, $c$ and $d$ be positive numbers such that $a+b+c+d=4$. Prove that: $a^3b+b^3c+c^3d+d^3a+4geq2(abc+abd+acd+bcd).$ It's true, I proved it.
$endgroup$
– Michael Rozenberg
Jan 21 at 3:41
$begingroup$
So there is misprint in the statement in the given inequality. It should read as $a_1^3a_2+a_2^3a_3+ldots+a_n^3a_1+ngeq 2sum^{n}_{i_1<i_2<ldots<i_{n-1}}a_{i_1}a_{i_2}ldots a_{i_{n-1}}$
$endgroup$
– Nikos Bagis
Jan 21 at 8:31