Mixing
to prove this inequality $sum a^3b+3ge 2(ab+bc+ca)$
$begingroup$
let $a,b,c>0$ and such $a+b+c=3$,show that
$$a^3b+b^3c+c^3a+3ge 2(ab+bc+ac)$$
This problem is from my question when $n=3$ case,I found not to prove it.
show this inequality with $sum_{i=1}^{n}a_{i}=n$
I have try use AM-GM since $$sum (a^3b+ab)ge 2sum a^2b$$
inequality substitution a.m.-g.m.-inequality buffalo-way
edited Jan 26 at 5:36
Michael Rozenberg
108k1895200
asked Jan 23 at 10:34
geromtygeromty
987523
$endgroup$
add a comment |
$begingroup$
let $a,b,c>0$ and such $a+b+c=3$,show that
$$a^3b+b^3c+c^3a+3ge 2(ab+bc+ac)$$
This problem is from my question when $n=3$ case,I found not to prove it.
show this inequality with $sum_{i=1}^{n}a_{i}=n$
I have try use AM-GM since $$sum (a^3b+ab)ge 2sum a^2b$$
inequality substitution a.m.-g.m.-inequality buffalo-way
edited Jan 26 at 5:36
Michael Rozenberg
108k1895200
asked Jan 23 at 10:34
geromtygeromty
987523
$endgroup$
add a comment |
$begingroup$
let $a,b,c>0$ and such $a+b+c=3$,show that
$$a^3b+b^3c+c^3a+3ge 2(ab+bc+ac)$$
This problem is from my question when $n=3$ case,I found not to prove it.
show this inequality with $sum_{i=1}^{n}a_{i}=n$
I have try use AM-GM since $$sum (a^3b+ab)ge 2sum a^2b$$
inequality substitution a.m.-g.m.-inequality buffalo-way
edited Jan 26 at 5:36
Michael Rozenberg
108k1895200
asked Jan 23 at 10:34
geromtygeromty
987523
$endgroup$
let $a,b,c>0$ and such $a+b+c=3$,show that
$$a^3b+b^3c+c^3a+3ge 2(ab+bc+ac)$$
This problem is from my question when $n=3$ case,I found not to prove it.
show this inequality with $sum_{i=1}^{n}a_{i}=n$
I have try use AM-GM since $$sum (a^3b+ab)ge 2sum a^2b$$
inequality substitution a.m.-g.m.-inequality buffalo-way
inequality substitution a.m.-g.m.-inequality buffalo-way
edited Jan 26 at 5:36
Michael Rozenberg
108k1895200
asked Jan 23 at 10:34
geromtygeromty
987523
edited Jan 26 at 5:36
Michael Rozenberg
108k1895200
asked Jan 23 at 10:34
geromtygeromty
987523
edited Jan 26 at 5:36
Michael Rozenberg
108k1895200
edited Jan 26 at 5:36
Michael Rozenberg
108k1895200
edited Jan 26 at 5:36
Michael Rozenberg
108k1895200
108k1895200
asked Jan 23 at 10:34
geromtygeromty
987523
asked Jan 23 at 10:34
geromtygeromty
987523
asked Jan 23 at 10:34
geromtygeromty
987523
987523
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We need to prove that
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4geq6(ab+ac+bc)(a+b+c)^2.$$
Let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.
Thus, by AM-GM
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4-6(ab+ac+bc)(a+b+c)^2=45(u^2-uv+v^2)a^2+$$
$$+9(3u^3+5u^2v-4uv^2+3v^3)a+u^4+25u^3v-6u^2v^2-2uv^3+v^4geq$$
$$geq25u^3v-8u^2v^2+frac{1}{2}v^4+2u^2v^2-2uv^3+frac{1}{2}v^4=$$
$$=frac{v}{2}left(25u^3+25u^3+v^3-16u^2vright)+v^2left(2u^2+frac{1}{2}v^2-2uvright)geq$$
$$geqfrac{v}{2}left(3sqrt[3]{left(25u^3right)^2v^3}-16u^2vright)+v^2left(2sqrt{2u^2cdotfrac{1}{2}v^2}-2uvright)=frac{u^2v^2}{2}left(3sqrt[3]{625}-16right)geq0.$$
Another proof.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2.$
Thus, $tgeq1$ and we need to prove that
$$sum_{cyc}2a^3b+6u^4geq12u^2v^2$$ or
$$sum_{cyc}(a^3b+a^3c)+6u^4-12u^2v^2geqsum_{cyc}(a^3c-a^3b)$$ or
$$27u^2v^2-18v^4-3uw^3+6u^4-12u^2v^2geq(a+b+c)(a-b)(b-c)(c-a)$$ or
$$2u^4+5u^2v^2-6v^4-uw^3geq u(a-b)(b-c)(c-a)$$ and since $$2u^4+5u^2v^2-6v^4-uw^3geq0,$$ it's enough to prove that
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq u^2(a-b)^2(b-c)^2(c-a)^2$$ or
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or
$$7u^2w^6+(26u^4-43u^2v^2+3v^4)uw^3+u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8geq0.$$
Now, since
$$u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8=v^8(t^4+5t^3-20t^2+12t+9)=$$
$$=v^8left(left(frac{t^2}{4}+2tright)(2t-3)^2+frac{1}{4}(7t^2-24t+36)right)geq0,$$
it's enough to prove our inequality for $26t^2-43t+3leq0,$ for which it's enough to prove that
$$(26t^2-43t+3)^2-28(t^4+5t^3-20t^2+12t+9)leq0$$ or
$$(t-1)^2(9+40t-24t^2)geq0,$$ which is true because
$$9+40t-24t^2=9+40t+frac{24}{26}left(-26t^2right)geq9+40t+frac{24}{26}left(3-43tright)>0.$$
Done!
answered Jan 23 at 10:51
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
1
$begingroup$
Is there a method that doesn't use this 'buffalo' approach?
$endgroup$
– user574848
Jan 23 at 10:57
$begingroup$
@user574848 I think, the proof by BW is an easiest here. There is also a proof by the $uvw$'s technique, but it's very ugly.
$endgroup$
– Michael Rozenberg
Jan 23 at 11:01
$begingroup$
You could also try the Lagrange multiplier method.
$endgroup$
– Klaus
Jan 23 at 11:05
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We need to prove that
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4geq6(ab+ac+bc)(a+b+c)^2.$$
Let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.
Thus, by AM-GM
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4-6(ab+ac+bc)(a+b+c)^2=45(u^2-uv+v^2)a^2+$$
$$+9(3u^3+5u^2v-4uv^2+3v^3)a+u^4+25u^3v-6u^2v^2-2uv^3+v^4geq$$
$$geq25u^3v-8u^2v^2+frac{1}{2}v^4+2u^2v^2-2uv^3+frac{1}{2}v^4=$$
$$=frac{v}{2}left(25u^3+25u^3+v^3-16u^2vright)+v^2left(2u^2+frac{1}{2}v^2-2uvright)geq$$
$$geqfrac{v}{2}left(3sqrt[3]{left(25u^3right)^2v^3}-16u^2vright)+v^2left(2sqrt{2u^2cdotfrac{1}{2}v^2}-2uvright)=frac{u^2v^2}{2}left(3sqrt[3]{625}-16right)geq0.$$
Another proof.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2.$
Thus, $tgeq1$ and we need to prove that
$$sum_{cyc}2a^3b+6u^4geq12u^2v^2$$ or
$$sum_{cyc}(a^3b+a^3c)+6u^4-12u^2v^2geqsum_{cyc}(a^3c-a^3b)$$ or
$$27u^2v^2-18v^4-3uw^3+6u^4-12u^2v^2geq(a+b+c)(a-b)(b-c)(c-a)$$ or
$$2u^4+5u^2v^2-6v^4-uw^3geq u(a-b)(b-c)(c-a)$$ and since $$2u^4+5u^2v^2-6v^4-uw^3geq0,$$ it's enough to prove that
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq u^2(a-b)^2(b-c)^2(c-a)^2$$ or
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or
$$7u^2w^6+(26u^4-43u^2v^2+3v^4)uw^3+u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8geq0.$$
Now, since
$$u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8=v^8(t^4+5t^3-20t^2+12t+9)=$$
$$=v^8left(left(frac{t^2}{4}+2tright)(2t-3)^2+frac{1}{4}(7t^2-24t+36)right)geq0,$$
it's enough to prove our inequality for $26t^2-43t+3leq0,$ for which it's enough to prove that
$$(26t^2-43t+3)^2-28(t^4+5t^3-20t^2+12t+9)leq0$$ or
$$(t-1)^2(9+40t-24t^2)geq0,$$ which is true because
$$9+40t-24t^2=9+40t+frac{24}{26}left(-26t^2right)geq9+40t+frac{24}{26}left(3-43tright)>0.$$
Done!
answered Jan 23 at 10:51
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
1
$begingroup$
Is there a method that doesn't use this 'buffalo' approach?
$endgroup$
– user574848
Jan 23 at 10:57
$begingroup$
@user574848 I think, the proof by BW is an easiest here. There is also a proof by the $uvw$'s technique, but it's very ugly.
$endgroup$
– Michael Rozenberg
Jan 23 at 11:01
$begingroup$
You could also try the Lagrange multiplier method.
$endgroup$
– Klaus
Jan 23 at 11:05
add a comment |
$begingroup$
We need to prove that
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4geq6(ab+ac+bc)(a+b+c)^2.$$
Let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.
Thus, by AM-GM
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4-6(ab+ac+bc)(a+b+c)^2=45(u^2-uv+v^2)a^2+$$
$$+9(3u^3+5u^2v-4uv^2+3v^3)a+u^4+25u^3v-6u^2v^2-2uv^3+v^4geq$$
$$geq25u^3v-8u^2v^2+frac{1}{2}v^4+2u^2v^2-2uv^3+frac{1}{2}v^4=$$
$$=frac{v}{2}left(25u^3+25u^3+v^3-16u^2vright)+v^2left(2u^2+frac{1}{2}v^2-2uvright)geq$$
$$geqfrac{v}{2}left(3sqrt[3]{left(25u^3right)^2v^3}-16u^2vright)+v^2left(2sqrt{2u^2cdotfrac{1}{2}v^2}-2uvright)=frac{u^2v^2}{2}left(3sqrt[3]{625}-16right)geq0.$$
Another proof.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2.$
Thus, $tgeq1$ and we need to prove that
$$sum_{cyc}2a^3b+6u^4geq12u^2v^2$$ or
$$sum_{cyc}(a^3b+a^3c)+6u^4-12u^2v^2geqsum_{cyc}(a^3c-a^3b)$$ or
$$27u^2v^2-18v^4-3uw^3+6u^4-12u^2v^2geq(a+b+c)(a-b)(b-c)(c-a)$$ or
$$2u^4+5u^2v^2-6v^4-uw^3geq u(a-b)(b-c)(c-a)$$ and since $$2u^4+5u^2v^2-6v^4-uw^3geq0,$$ it's enough to prove that
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq u^2(a-b)^2(b-c)^2(c-a)^2$$ or
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or
$$7u^2w^6+(26u^4-43u^2v^2+3v^4)uw^3+u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8geq0.$$
Now, since
$$u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8=v^8(t^4+5t^3-20t^2+12t+9)=$$
$$=v^8left(left(frac{t^2}{4}+2tright)(2t-3)^2+frac{1}{4}(7t^2-24t+36)right)geq0,$$
it's enough to prove our inequality for $26t^2-43t+3leq0,$ for which it's enough to prove that
$$(26t^2-43t+3)^2-28(t^4+5t^3-20t^2+12t+9)leq0$$ or
$$(t-1)^2(9+40t-24t^2)geq0,$$ which is true because
$$9+40t-24t^2=9+40t+frac{24}{26}left(-26t^2right)geq9+40t+frac{24}{26}left(3-43tright)>0.$$
Done!
answered Jan 23 at 10:51
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
1
$begingroup$
Is there a method that doesn't use this 'buffalo' approach?
$endgroup$
– user574848
Jan 23 at 10:57
$begingroup$
@user574848 I think, the proof by BW is an easiest here. There is also a proof by the $uvw$'s technique, but it's very ugly.
$endgroup$
– Michael Rozenberg
Jan 23 at 11:01
$begingroup$
You could also try the Lagrange multiplier method.
$endgroup$
– Klaus
Jan 23 at 11:05
add a comment |
$begingroup$
We need to prove that
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4geq6(ab+ac+bc)(a+b+c)^2.$$
Let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.
Thus, by AM-GM
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4-6(ab+ac+bc)(a+b+c)^2=45(u^2-uv+v^2)a^2+$$
$$+9(3u^3+5u^2v-4uv^2+3v^3)a+u^4+25u^3v-6u^2v^2-2uv^3+v^4geq$$
$$geq25u^3v-8u^2v^2+frac{1}{2}v^4+2u^2v^2-2uv^3+frac{1}{2}v^4=$$
$$=frac{v}{2}left(25u^3+25u^3+v^3-16u^2vright)+v^2left(2u^2+frac{1}{2}v^2-2uvright)geq$$
$$geqfrac{v}{2}left(3sqrt[3]{left(25u^3right)^2v^3}-16u^2vright)+v^2left(2sqrt{2u^2cdotfrac{1}{2}v^2}-2uvright)=frac{u^2v^2}{2}left(3sqrt[3]{625}-16right)geq0.$$
Another proof.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2.$
Thus, $tgeq1$ and we need to prove that
$$sum_{cyc}2a^3b+6u^4geq12u^2v^2$$ or
$$sum_{cyc}(a^3b+a^3c)+6u^4-12u^2v^2geqsum_{cyc}(a^3c-a^3b)$$ or
$$27u^2v^2-18v^4-3uw^3+6u^4-12u^2v^2geq(a+b+c)(a-b)(b-c)(c-a)$$ or
$$2u^4+5u^2v^2-6v^4-uw^3geq u(a-b)(b-c)(c-a)$$ and since $$2u^4+5u^2v^2-6v^4-uw^3geq0,$$ it's enough to prove that
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq u^2(a-b)^2(b-c)^2(c-a)^2$$ or
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or
$$7u^2w^6+(26u^4-43u^2v^2+3v^4)uw^3+u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8geq0.$$
Now, since
$$u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8=v^8(t^4+5t^3-20t^2+12t+9)=$$
$$=v^8left(left(frac{t^2}{4}+2tright)(2t-3)^2+frac{1}{4}(7t^2-24t+36)right)geq0,$$
it's enough to prove our inequality for $26t^2-43t+3leq0,$ for which it's enough to prove that
$$(26t^2-43t+3)^2-28(t^4+5t^3-20t^2+12t+9)leq0$$ or
$$(t-1)^2(9+40t-24t^2)geq0,$$ which is true because
$$9+40t-24t^2=9+40t+frac{24}{26}left(-26t^2right)geq9+40t+frac{24}{26}left(3-43tright)>0.$$
Done!
answered Jan 23 at 10:51
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
We need to prove that
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4geq6(ab+ac+bc)(a+b+c)^2.$$
Let $a=min{a,b,c}$, $b=a+u$ and $c=a+v$.
Thus, by AM-GM
$$27(a^3b+b^3c+c^3a)+(a+b+c)^4-6(ab+ac+bc)(a+b+c)^2=45(u^2-uv+v^2)a^2+$$
$$+9(3u^3+5u^2v-4uv^2+3v^3)a+u^4+25u^3v-6u^2v^2-2uv^3+v^4geq$$
$$geq25u^3v-8u^2v^2+frac{1}{2}v^4+2u^2v^2-2uv^3+frac{1}{2}v^4=$$
$$=frac{v}{2}left(25u^3+25u^3+v^3-16u^2vright)+v^2left(2u^2+frac{1}{2}v^2-2uvright)geq$$
$$geqfrac{v}{2}left(3sqrt[3]{left(25u^3right)^2v^3}-16u^2vright)+v^2left(2sqrt{2u^2cdotfrac{1}{2}v^2}-2uvright)=frac{u^2v^2}{2}left(3sqrt[3]{625}-16right)geq0.$$
Another proof.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2.$
Thus, $tgeq1$ and we need to prove that
$$sum_{cyc}2a^3b+6u^4geq12u^2v^2$$ or
$$sum_{cyc}(a^3b+a^3c)+6u^4-12u^2v^2geqsum_{cyc}(a^3c-a^3b)$$ or
$$27u^2v^2-18v^4-3uw^3+6u^4-12u^2v^2geq(a+b+c)(a-b)(b-c)(c-a)$$ or
$$2u^4+5u^2v^2-6v^4-uw^3geq u(a-b)(b-c)(c-a)$$ and since $$2u^4+5u^2v^2-6v^4-uw^3geq0,$$ it's enough to prove that
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq u^2(a-b)^2(b-c)^2(c-a)^2$$ or
$$(2u^4+5u^2v^2-6v^4-uw^3)^2geq27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or
$$7u^2w^6+(26u^4-43u^2v^2+3v^4)uw^3+u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8geq0.$$
Now, since
$$u^8+5u^6v^2-20u^4v^4+12u^2v^6+9v^8=v^8(t^4+5t^3-20t^2+12t+9)=$$
$$=v^8left(left(frac{t^2}{4}+2tright)(2t-3)^2+frac{1}{4}(7t^2-24t+36)right)geq0,$$
it's enough to prove our inequality for $26t^2-43t+3leq0,$ for which it's enough to prove that
$$(26t^2-43t+3)^2-28(t^4+5t^3-20t^2+12t+9)leq0$$ or
$$(t-1)^2(9+40t-24t^2)geq0,$$ which is true because
$$9+40t-24t^2=9+40t+frac{24}{26}left(-26t^2right)geq9+40t+frac{24}{26}left(3-43tright)>0.$$
Done!
answered Jan 23 at 10:51
Michael RozenbergMichael Rozenberg
108k1895200
edited Jan 23 at 12:50
answered Jan 23 at 10:51
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 10:51
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 10:51
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
1
$begingroup$
Is there a method that doesn't use this 'buffalo' approach?
$endgroup$
– user574848
Jan 23 at 10:57
$begingroup$
@user574848 I think, the proof by BW is an easiest here. There is also a proof by the $uvw$'s technique, but it's very ugly.
$endgroup$
– Michael Rozenberg
Jan 23 at 11:01
$begingroup$
You could also try the Lagrange multiplier method.
$endgroup$
– Klaus
Jan 23 at 11:05
add a comment |
1
$begingroup$
Is there a method that doesn't use this 'buffalo' approach?
$endgroup$
– user574848
Jan 23 at 10:57
$begingroup$
@user574848 I think, the proof by BW is an easiest here. There is also a proof by the $uvw$'s technique, but it's very ugly.
$endgroup$
– Michael Rozenberg
Jan 23 at 11:01
$begingroup$
You could also try the Lagrange multiplier method.
$endgroup$
– Klaus
Jan 23 at 11:05
1
1
$begingroup$
Is there a method that doesn't use this 'buffalo' approach?
$endgroup$
– user574848
Jan 23 at 10:57
$begingroup$
Is there a method that doesn't use this 'buffalo' approach?
$endgroup$
– user574848
Jan 23 at 10:57
$begingroup$
@user574848 I think, the proof by BW is an easiest here. There is also a proof by the $uvw$'s technique, but it's very ugly.
$endgroup$
– Michael Rozenberg
Jan 23 at 11:01
$begingroup$
@user574848 I think, the proof by BW is an easiest here. There is also a proof by the $uvw$'s technique, but it's very ugly.
$endgroup$
– Michael Rozenberg
Jan 23 at 11:01
$begingroup$
You could also try the Lagrange multiplier method.
$endgroup$
– Klaus
Jan 23 at 11:05
$begingroup$
You could also try the Lagrange multiplier method.
$endgroup$
– Klaus
Jan 23 at 11:05
add a comment |
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