Mixing
$frac{|a|}{|b-c|} + frac{|b|}{|c-a|} + frac{|c|}{|b-a|} geq 2$
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If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=frac{|a|}{|b-c|} + frac{|b|}{|c-a|} + frac{|c|}{|b-a|} geq 2.$$
Using inequality $ |x-y|leq |x|+|y|$ we showed that $ S >frac{2}{3}.$
For $b = 2a, c = 3a, S=5,$ that is, the sum has values greater than 2, but I have not been able to prove this.
inequality
asked yesterday
medicu
3,016813
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up vote
5
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favorite
If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=frac{|a|}{|b-c|} + frac{|b|}{|c-a|} + frac{|c|}{|b-a|} geq 2.$$
Using inequality $ |x-y|leq |x|+|y|$ we showed that $ S >frac{2}{3}.$
For $b = 2a, c = 3a, S=5,$ that is, the sum has values greater than 2, but I have not been able to prove this.
inequality
asked yesterday
medicu
3,016813
Put everyting on one side, multiply everything by $|b-c||c-a||b-a|, and go from there.
– user3482749
yesterday
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up vote
5
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favorite
up vote
5
down vote
favorite
If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=frac{|a|}{|b-c|} + frac{|b|}{|c-a|} + frac{|c|}{|b-a|} geq 2.$$
Using inequality $ |x-y|leq |x|+|y|$ we showed that $ S >frac{2}{3}.$
For $b = 2a, c = 3a, S=5,$ that is, the sum has values greater than 2, but I have not been able to prove this.
inequality
asked yesterday
medicu
3,016813
If $a, b, c$ are distinct real numbers then you demonstrate that:
$$ S=frac{|a|}{|b-c|} + frac{|b|}{|c-a|} + frac{|c|}{|b-a|} geq 2.$$
Using inequality $ |x-y|leq |x|+|y|$ we showed that $ S >frac{2}{3}.$
For $b = 2a, c = 3a, S=5,$ that is, the sum has values greater than 2, but I have not been able to prove this.
inequality
inequality
asked yesterday
medicu
3,016813
asked yesterday
medicu
3,016813
asked yesterday
medicu
3,016813
asked yesterday
medicu
3,016813
asked yesterday
medicu
3,016813
3,016813
Put everyting on one side, multiply everything by $|b-c||c-a||b-a|, and go from there.
– user3482749
yesterday
add a comment |
Put everyting on one side, multiply everything by $|b-c||c-a||b-a|, and go from there.
– user3482749
yesterday
Put everyting on one side, multiply everything by $|b-c||c-a||b-a|, and go from there.
– user3482749
yesterday
Put everyting on one side, multiply everything by $|b-c||c-a||b-a|, and go from there.
– user3482749
yesterday
add a comment |
4 Answers
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4
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accepted
Let $frac{a}{b-c}=x$, $frac{b}{c-a}=y$ and $frac{c}{a-b}=z$.
Thus, $$xy+xz+yz=sum_{cyc}frac{ab}{(b-c)(c-a)}=frac{sumlimits_{cyc}ab(a-b)}{(a-b)(b-c)(c-a)}=frac{(a-b)(a-c)(b-c)}{(a-b)(b-c)(c-a)}=-1.$$
Id est,
$$sum_{cyc}left|frac{a}{b-c}right|=sqrt{left(|x|+|y|+|z|right)^2}=sqrt{x^2+y^2+z^2+2sumlimits_{cyc}|xy|}=$$
$$=sqrt{(x+y+z)^2+2+2sumlimits_{cyc}|xy|}geqsqrt{2+2}=2.$$
answered yesterday
Michael Rozenberg
94.2k1588183
1
+1, nice answer.
– the_candyman
yesterday
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up vote
3
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Without loss of generality we can assume that $b$ and $c$ have the same sign. Then $|b|=|b-c|+|c|$. Also, $|c-a|le |c|+|a|$ and $|a-b| le |a|+|b-c|+|c|$. Therefore
$$
frac{|a|}{|b-c|}+frac{|b|}{|c-a|}+frac{|c|}{|a-b|} ge frac{|a|}{|b-c|} + frac{|b-c|+|c|}{|c|+|a|} + frac{|c|}{|a|+|b-c|+|c|} = (*).$$
Denote $x=|a|$, $y=|b-c|$, $z=|c|$. Then, by Schwarz and by $x^2+y^2 ge 2xy$, we get
begin{align*}
(*) & = frac{x}{y}+frac{y+z}{z+x}+frac{z}{x+y+z} \
&ge frac{(x+(y+z)+z)^2}{xy+(y+z)(z+x)+z(x+y+z)} \
&= frac{x^2+y^2+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&ge frac{2xy+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&=2.end{align*}
answered yesterday
timon92
4,0691725
+1, nice answer.
– the_candyman
yesterday
add a comment |
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0
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Notice that $|x-y|leq |x|+|y|$ implies that:
$$frac{|z|}{|x-y|} geq frac{|z|}{|x|+|y|}.$$
Then:
$$S geq frac{|a|}{|b|+|c|} + frac{|b|}{|a|+|c|} + frac{|c|}{|a|+|b|} = \
=frac{A}{B+C} + frac{B}{A+C} + frac{C}{A+B},$$
where I set $A = |a|, B = |b|$ and $C= |c|$ for the sake of simplicity.
Multiplying and dividing numerators and denominators by $(B+C)(A+C)(A+B)$, we get that:
$$S geq frac{A^3+A^2B + A^2 C + AB^2 + 3ABC + AC^2 + B^3 + B^2C + BC^2 + C^3 }{(B+C)(A+C)(A+B)}.$$
Notice that the denominator can be expanded as follows:
$$D = (B+C)(A+C)(A+B) = A^2B + A^2C + AB^2 + 2ABC + AC^2 + B^2C + BC^2.$$
Therefore, the numerator can be rewritten as:
$$ N= D + A^3 +B^3 + C^3 + ABC.$$
In other words:
$$S geq frac{D + A^3 +B^3 + C^3 + ABC}{D} > frac{D}{D} = 1.$$
answered yesterday
the_candyman
8,54921944
How does this prove that $S ge 2$?
– timon92
yesterday
@timon92 It doesn't. This is my attempt. If I can find a better lower bound, then I will post it.
– the_candyman
yesterday
1
Fair enough. Also, using the so-called Nesbitt inequality one can get better bound: $Sge frac{|a|}{|b|+|c|}+frac{|b|}{|c|+|a|}+frac{|c|}{|a|+|b|} ge frac 32.$
– timon92
yesterday
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0
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I tried to give a simple proof by reducing to simpler and simpler cases, all the time doing "nothing", and the final rephrasing is simple, too.
First of all, let us note that the (LHS of the) given inequality is invariated by permutations of the letters / variables $a,b,c$. So we may and do assume $ale ble c$.
Let $s,t>0$ be then such that $b=a+s$, $c=b+t=a+s+t$.
Replacing the initial $a,b,c$ with $-c,-b,-a$, we may and do assume $bge 0$.
The given inequality can now be rewritten equivalently:
$$
S=
frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
ge 2 .
$$If $a$ is $ge0$, then all three numbers $a,b,c$ are $ge 0$, and we can write
$$
begin{aligned}
S
&= frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
\
&ge
frac{0}t +
frac{s}{s+t} +
frac{s+t}s
%ge 2sqrt{
%frac{s}{s+t} cdot
%frac{s+t}s}
ge2 ,qquad(age 0)
end{aligned}
$$
and the last $ge$ (AM-GM) is moreover $>0$ (because $tne 0$). The case with $a < b < c=a+s+tle 0$ is similar.So the single interesting case is the one with $a<0le b<c$ . We rewrite the inequality in the form:
$$
frac{|b-s|}t+frac b{s+t}+frac{b+t}sge 2 .
$$
Here, $b$ is allowed to vary between $0$ and $s$, and the first modulus is $|a|=-a=-(b-s)=s-b$, so we have to show:
$$
frac{s-b}t+frac b{s+t}+frac{b+t}sge 2 .
$$
But this is a linear (ergo convex) function in $b$, so the inequality has to be tested only for the possible extremities, $b=0$ and $b=s$. For $b=0$ we obtain $frac st+frac tsge 2$, true, with equality for $s=t$. For $b=s$ we get $frac s{s+t}+frac {s+t}sge 2$, an inequality of the same shape, but we cannot get equality. (For intermediate values of $b$ we get intermediate values of the expression.)
answered 10 hours ago
dan_fulea
5,8201312
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4 Answers
4
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oldest
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4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let $frac{a}{b-c}=x$, $frac{b}{c-a}=y$ and $frac{c}{a-b}=z$.
Thus, $$xy+xz+yz=sum_{cyc}frac{ab}{(b-c)(c-a)}=frac{sumlimits_{cyc}ab(a-b)}{(a-b)(b-c)(c-a)}=frac{(a-b)(a-c)(b-c)}{(a-b)(b-c)(c-a)}=-1.$$
Id est,
$$sum_{cyc}left|frac{a}{b-c}right|=sqrt{left(|x|+|y|+|z|right)^2}=sqrt{x^2+y^2+z^2+2sumlimits_{cyc}|xy|}=$$
$$=sqrt{(x+y+z)^2+2+2sumlimits_{cyc}|xy|}geqsqrt{2+2}=2.$$
answered yesterday
Michael Rozenberg
94.2k1588183
1
+1, nice answer.
– the_candyman
yesterday
add a comment |
up vote
4
down vote
accepted
Let $frac{a}{b-c}=x$, $frac{b}{c-a}=y$ and $frac{c}{a-b}=z$.
Thus, $$xy+xz+yz=sum_{cyc}frac{ab}{(b-c)(c-a)}=frac{sumlimits_{cyc}ab(a-b)}{(a-b)(b-c)(c-a)}=frac{(a-b)(a-c)(b-c)}{(a-b)(b-c)(c-a)}=-1.$$
Id est,
$$sum_{cyc}left|frac{a}{b-c}right|=sqrt{left(|x|+|y|+|z|right)^2}=sqrt{x^2+y^2+z^2+2sumlimits_{cyc}|xy|}=$$
$$=sqrt{(x+y+z)^2+2+2sumlimits_{cyc}|xy|}geqsqrt{2+2}=2.$$
answered yesterday
Michael Rozenberg
94.2k1588183
1
+1, nice answer.
– the_candyman
yesterday
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let $frac{a}{b-c}=x$, $frac{b}{c-a}=y$ and $frac{c}{a-b}=z$.
Thus, $$xy+xz+yz=sum_{cyc}frac{ab}{(b-c)(c-a)}=frac{sumlimits_{cyc}ab(a-b)}{(a-b)(b-c)(c-a)}=frac{(a-b)(a-c)(b-c)}{(a-b)(b-c)(c-a)}=-1.$$
Id est,
$$sum_{cyc}left|frac{a}{b-c}right|=sqrt{left(|x|+|y|+|z|right)^2}=sqrt{x^2+y^2+z^2+2sumlimits_{cyc}|xy|}=$$
$$=sqrt{(x+y+z)^2+2+2sumlimits_{cyc}|xy|}geqsqrt{2+2}=2.$$
answered yesterday
Michael Rozenberg
94.2k1588183
Let $frac{a}{b-c}=x$, $frac{b}{c-a}=y$ and $frac{c}{a-b}=z$.
Thus, $$xy+xz+yz=sum_{cyc}frac{ab}{(b-c)(c-a)}=frac{sumlimits_{cyc}ab(a-b)}{(a-b)(b-c)(c-a)}=frac{(a-b)(a-c)(b-c)}{(a-b)(b-c)(c-a)}=-1.$$
Id est,
$$sum_{cyc}left|frac{a}{b-c}right|=sqrt{left(|x|+|y|+|z|right)^2}=sqrt{x^2+y^2+z^2+2sumlimits_{cyc}|xy|}=$$
$$=sqrt{(x+y+z)^2+2+2sumlimits_{cyc}|xy|}geqsqrt{2+2}=2.$$
answered yesterday
Michael Rozenberg
94.2k1588183
answered yesterday
Michael Rozenberg
94.2k1588183
answered yesterday
Michael Rozenberg
94.2k1588183
answered yesterday
Michael Rozenberg
94.2k1588183
94.2k1588183
1
+1, nice answer.
– the_candyman
yesterday
add a comment |
1
+1, nice answer.
– the_candyman
yesterday
1
1
+1, nice answer.
– the_candyman
yesterday
+1, nice answer.
– the_candyman
yesterday
add a comment |
up vote
3
down vote
Without loss of generality we can assume that $b$ and $c$ have the same sign. Then $|b|=|b-c|+|c|$. Also, $|c-a|le |c|+|a|$ and $|a-b| le |a|+|b-c|+|c|$. Therefore
$$
frac{|a|}{|b-c|}+frac{|b|}{|c-a|}+frac{|c|}{|a-b|} ge frac{|a|}{|b-c|} + frac{|b-c|+|c|}{|c|+|a|} + frac{|c|}{|a|+|b-c|+|c|} = (*).$$
Denote $x=|a|$, $y=|b-c|$, $z=|c|$. Then, by Schwarz and by $x^2+y^2 ge 2xy$, we get
begin{align*}
(*) & = frac{x}{y}+frac{y+z}{z+x}+frac{z}{x+y+z} \
&ge frac{(x+(y+z)+z)^2}{xy+(y+z)(z+x)+z(x+y+z)} \
&= frac{x^2+y^2+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&ge frac{2xy+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&=2.end{align*}
answered yesterday
timon92
4,0691725
+1, nice answer.
– the_candyman
yesterday
add a comment |
up vote
3
down vote
Without loss of generality we can assume that $b$ and $c$ have the same sign. Then $|b|=|b-c|+|c|$. Also, $|c-a|le |c|+|a|$ and $|a-b| le |a|+|b-c|+|c|$. Therefore
$$
frac{|a|}{|b-c|}+frac{|b|}{|c-a|}+frac{|c|}{|a-b|} ge frac{|a|}{|b-c|} + frac{|b-c|+|c|}{|c|+|a|} + frac{|c|}{|a|+|b-c|+|c|} = (*).$$
Denote $x=|a|$, $y=|b-c|$, $z=|c|$. Then, by Schwarz and by $x^2+y^2 ge 2xy$, we get
begin{align*}
(*) & = frac{x}{y}+frac{y+z}{z+x}+frac{z}{x+y+z} \
&ge frac{(x+(y+z)+z)^2}{xy+(y+z)(z+x)+z(x+y+z)} \
&= frac{x^2+y^2+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&ge frac{2xy+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&=2.end{align*}
answered yesterday
timon92
4,0691725
+1, nice answer.
– the_candyman
yesterday
add a comment |
up vote
3
down vote
up vote
3
down vote
Without loss of generality we can assume that $b$ and $c$ have the same sign. Then $|b|=|b-c|+|c|$. Also, $|c-a|le |c|+|a|$ and $|a-b| le |a|+|b-c|+|c|$. Therefore
$$
frac{|a|}{|b-c|}+frac{|b|}{|c-a|}+frac{|c|}{|a-b|} ge frac{|a|}{|b-c|} + frac{|b-c|+|c|}{|c|+|a|} + frac{|c|}{|a|+|b-c|+|c|} = (*).$$
Denote $x=|a|$, $y=|b-c|$, $z=|c|$. Then, by Schwarz and by $x^2+y^2 ge 2xy$, we get
begin{align*}
(*) & = frac{x}{y}+frac{y+z}{z+x}+frac{z}{x+y+z} \
&ge frac{(x+(y+z)+z)^2}{xy+(y+z)(z+x)+z(x+y+z)} \
&= frac{x^2+y^2+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&ge frac{2xy+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&=2.end{align*}
answered yesterday
timon92
4,0691725
Without loss of generality we can assume that $b$ and $c$ have the same sign. Then $|b|=|b-c|+|c|$. Also, $|c-a|le |c|+|a|$ and $|a-b| le |a|+|b-c|+|c|$. Therefore
$$
frac{|a|}{|b-c|}+frac{|b|}{|c-a|}+frac{|c|}{|a-b|} ge frac{|a|}{|b-c|} + frac{|b-c|+|c|}{|c|+|a|} + frac{|c|}{|a|+|b-c|+|c|} = (*).$$
Denote $x=|a|$, $y=|b-c|$, $z=|c|$. Then, by Schwarz and by $x^2+y^2 ge 2xy$, we get
begin{align*}
(*) & = frac{x}{y}+frac{y+z}{z+x}+frac{z}{x+y+z} \
&ge frac{(x+(y+z)+z)^2}{xy+(y+z)(z+x)+z(x+y+z)} \
&= frac{x^2+y^2+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&ge frac{2xy+4z^2+2xy+4yz+4zx}{2z^2+2xy+2yz+2zx} \
&=2.end{align*}
answered yesterday
timon92
4,0691725
answered yesterday
timon92
4,0691725
answered yesterday
timon92
4,0691725
answered yesterday
timon92
4,0691725
4,0691725
+1, nice answer.
– the_candyman
yesterday
add a comment |
+1, nice answer.
– the_candyman
yesterday
+1, nice answer.
– the_candyman
yesterday
+1, nice answer.
– the_candyman
yesterday
add a comment |
up vote
0
down vote
Notice that $|x-y|leq |x|+|y|$ implies that:
$$frac{|z|}{|x-y|} geq frac{|z|}{|x|+|y|}.$$
Then:
$$S geq frac{|a|}{|b|+|c|} + frac{|b|}{|a|+|c|} + frac{|c|}{|a|+|b|} = \
=frac{A}{B+C} + frac{B}{A+C} + frac{C}{A+B},$$
where I set $A = |a|, B = |b|$ and $C= |c|$ for the sake of simplicity.
Multiplying and dividing numerators and denominators by $(B+C)(A+C)(A+B)$, we get that:
$$S geq frac{A^3+A^2B + A^2 C + AB^2 + 3ABC + AC^2 + B^3 + B^2C + BC^2 + C^3 }{(B+C)(A+C)(A+B)}.$$
Notice that the denominator can be expanded as follows:
$$D = (B+C)(A+C)(A+B) = A^2B + A^2C + AB^2 + 2ABC + AC^2 + B^2C + BC^2.$$
Therefore, the numerator can be rewritten as:
$$ N= D + A^3 +B^3 + C^3 + ABC.$$
In other words:
$$S geq frac{D + A^3 +B^3 + C^3 + ABC}{D} > frac{D}{D} = 1.$$
answered yesterday
the_candyman
8,54921944
How does this prove that $S ge 2$?
– timon92
yesterday
@timon92 It doesn't. This is my attempt. If I can find a better lower bound, then I will post it.
– the_candyman
yesterday
1
Fair enough. Also, using the so-called Nesbitt inequality one can get better bound: $Sge frac{|a|}{|b|+|c|}+frac{|b|}{|c|+|a|}+frac{|c|}{|a|+|b|} ge frac 32.$
– timon92
yesterday
add a comment |
up vote
0
down vote
Notice that $|x-y|leq |x|+|y|$ implies that:
$$frac{|z|}{|x-y|} geq frac{|z|}{|x|+|y|}.$$
Then:
$$S geq frac{|a|}{|b|+|c|} + frac{|b|}{|a|+|c|} + frac{|c|}{|a|+|b|} = \
=frac{A}{B+C} + frac{B}{A+C} + frac{C}{A+B},$$
where I set $A = |a|, B = |b|$ and $C= |c|$ for the sake of simplicity.
Multiplying and dividing numerators and denominators by $(B+C)(A+C)(A+B)$, we get that:
$$S geq frac{A^3+A^2B + A^2 C + AB^2 + 3ABC + AC^2 + B^3 + B^2C + BC^2 + C^3 }{(B+C)(A+C)(A+B)}.$$
Notice that the denominator can be expanded as follows:
$$D = (B+C)(A+C)(A+B) = A^2B + A^2C + AB^2 + 2ABC + AC^2 + B^2C + BC^2.$$
Therefore, the numerator can be rewritten as:
$$ N= D + A^3 +B^3 + C^3 + ABC.$$
In other words:
$$S geq frac{D + A^3 +B^3 + C^3 + ABC}{D} > frac{D}{D} = 1.$$
answered yesterday
the_candyman
8,54921944
How does this prove that $S ge 2$?
– timon92
yesterday
@timon92 It doesn't. This is my attempt. If I can find a better lower bound, then I will post it.
– the_candyman
yesterday
1
Fair enough. Also, using the so-called Nesbitt inequality one can get better bound: $Sge frac{|a|}{|b|+|c|}+frac{|b|}{|c|+|a|}+frac{|c|}{|a|+|b|} ge frac 32.$
– timon92
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
Notice that $|x-y|leq |x|+|y|$ implies that:
$$frac{|z|}{|x-y|} geq frac{|z|}{|x|+|y|}.$$
Then:
$$S geq frac{|a|}{|b|+|c|} + frac{|b|}{|a|+|c|} + frac{|c|}{|a|+|b|} = \
=frac{A}{B+C} + frac{B}{A+C} + frac{C}{A+B},$$
where I set $A = |a|, B = |b|$ and $C= |c|$ for the sake of simplicity.
Multiplying and dividing numerators and denominators by $(B+C)(A+C)(A+B)$, we get that:
$$S geq frac{A^3+A^2B + A^2 C + AB^2 + 3ABC + AC^2 + B^3 + B^2C + BC^2 + C^3 }{(B+C)(A+C)(A+B)}.$$
Notice that the denominator can be expanded as follows:
$$D = (B+C)(A+C)(A+B) = A^2B + A^2C + AB^2 + 2ABC + AC^2 + B^2C + BC^2.$$
Therefore, the numerator can be rewritten as:
$$ N= D + A^3 +B^3 + C^3 + ABC.$$
In other words:
$$S geq frac{D + A^3 +B^3 + C^3 + ABC}{D} > frac{D}{D} = 1.$$
answered yesterday
the_candyman
8,54921944
Notice that $|x-y|leq |x|+|y|$ implies that:
$$frac{|z|}{|x-y|} geq frac{|z|}{|x|+|y|}.$$
Then:
$$S geq frac{|a|}{|b|+|c|} + frac{|b|}{|a|+|c|} + frac{|c|}{|a|+|b|} = \
=frac{A}{B+C} + frac{B}{A+C} + frac{C}{A+B},$$
where I set $A = |a|, B = |b|$ and $C= |c|$ for the sake of simplicity.
Multiplying and dividing numerators and denominators by $(B+C)(A+C)(A+B)$, we get that:
$$S geq frac{A^3+A^2B + A^2 C + AB^2 + 3ABC + AC^2 + B^3 + B^2C + BC^2 + C^3 }{(B+C)(A+C)(A+B)}.$$
Notice that the denominator can be expanded as follows:
$$D = (B+C)(A+C)(A+B) = A^2B + A^2C + AB^2 + 2ABC + AC^2 + B^2C + BC^2.$$
Therefore, the numerator can be rewritten as:
$$ N= D + A^3 +B^3 + C^3 + ABC.$$
In other words:
$$S geq frac{D + A^3 +B^3 + C^3 + ABC}{D} > frac{D}{D} = 1.$$
answered yesterday
the_candyman
8,54921944
edited yesterday
answered yesterday
the_candyman
8,54921944
answered yesterday
the_candyman
8,54921944
answered yesterday
the_candyman
8,54921944
8,54921944
How does this prove that $S ge 2$?
– timon92
yesterday
@timon92 It doesn't. This is my attempt. If I can find a better lower bound, then I will post it.
– the_candyman
yesterday
1
Fair enough. Also, using the so-called Nesbitt inequality one can get better bound: $Sge frac{|a|}{|b|+|c|}+frac{|b|}{|c|+|a|}+frac{|c|}{|a|+|b|} ge frac 32.$
– timon92
yesterday
add a comment |
How does this prove that $S ge 2$?
– timon92
yesterday
@timon92 It doesn't. This is my attempt. If I can find a better lower bound, then I will post it.
– the_candyman
yesterday
1
Fair enough. Also, using the so-called Nesbitt inequality one can get better bound: $Sge frac{|a|}{|b|+|c|}+frac{|b|}{|c|+|a|}+frac{|c|}{|a|+|b|} ge frac 32.$
– timon92
yesterday
How does this prove that $S ge 2$?
– timon92
yesterday
How does this prove that $S ge 2$?
– timon92
yesterday
@timon92 It doesn't. This is my attempt. If I can find a better lower bound, then I will post it.
– the_candyman
yesterday
@timon92 It doesn't. This is my attempt. If I can find a better lower bound, then I will post it.
– the_candyman
yesterday
1
1
Fair enough. Also, using the so-called Nesbitt inequality one can get better bound: $Sge frac{|a|}{|b|+|c|}+frac{|b|}{|c|+|a|}+frac{|c|}{|a|+|b|} ge frac 32.$
– timon92
yesterday
Fair enough. Also, using the so-called Nesbitt inequality one can get better bound: $Sge frac{|a|}{|b|+|c|}+frac{|b|}{|c|+|a|}+frac{|c|}{|a|+|b|} ge frac 32.$
– timon92
yesterday
add a comment |
up vote
0
down vote
I tried to give a simple proof by reducing to simpler and simpler cases, all the time doing "nothing", and the final rephrasing is simple, too.
First of all, let us note that the (LHS of the) given inequality is invariated by permutations of the letters / variables $a,b,c$. So we may and do assume $ale ble c$.
Let $s,t>0$ be then such that $b=a+s$, $c=b+t=a+s+t$.
Replacing the initial $a,b,c$ with $-c,-b,-a$, we may and do assume $bge 0$.
The given inequality can now be rewritten equivalently:
$$
S=
frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
ge 2 .
$$If $a$ is $ge0$, then all three numbers $a,b,c$ are $ge 0$, and we can write
$$
begin{aligned}
S
&= frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
\
&ge
frac{0}t +
frac{s}{s+t} +
frac{s+t}s
%ge 2sqrt{
%frac{s}{s+t} cdot
%frac{s+t}s}
ge2 ,qquad(age 0)
end{aligned}
$$
and the last $ge$ (AM-GM) is moreover $>0$ (because $tne 0$). The case with $a < b < c=a+s+tle 0$ is similar.So the single interesting case is the one with $a<0le b<c$ . We rewrite the inequality in the form:
$$
frac{|b-s|}t+frac b{s+t}+frac{b+t}sge 2 .
$$
Here, $b$ is allowed to vary between $0$ and $s$, and the first modulus is $|a|=-a=-(b-s)=s-b$, so we have to show:
$$
frac{s-b}t+frac b{s+t}+frac{b+t}sge 2 .
$$
But this is a linear (ergo convex) function in $b$, so the inequality has to be tested only for the possible extremities, $b=0$ and $b=s$. For $b=0$ we obtain $frac st+frac tsge 2$, true, with equality for $s=t$. For $b=s$ we get $frac s{s+t}+frac {s+t}sge 2$, an inequality of the same shape, but we cannot get equality. (For intermediate values of $b$ we get intermediate values of the expression.)
answered 10 hours ago
dan_fulea
5,8201312
add a comment |
up vote
0
down vote
I tried to give a simple proof by reducing to simpler and simpler cases, all the time doing "nothing", and the final rephrasing is simple, too.
First of all, let us note that the (LHS of the) given inequality is invariated by permutations of the letters / variables $a,b,c$. So we may and do assume $ale ble c$.
Let $s,t>0$ be then such that $b=a+s$, $c=b+t=a+s+t$.
Replacing the initial $a,b,c$ with $-c,-b,-a$, we may and do assume $bge 0$.
The given inequality can now be rewritten equivalently:
$$
S=
frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
ge 2 .
$$If $a$ is $ge0$, then all three numbers $a,b,c$ are $ge 0$, and we can write
$$
begin{aligned}
S
&= frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
\
&ge
frac{0}t +
frac{s}{s+t} +
frac{s+t}s
%ge 2sqrt{
%frac{s}{s+t} cdot
%frac{s+t}s}
ge2 ,qquad(age 0)
end{aligned}
$$
and the last $ge$ (AM-GM) is moreover $>0$ (because $tne 0$). The case with $a < b < c=a+s+tle 0$ is similar.So the single interesting case is the one with $a<0le b<c$ . We rewrite the inequality in the form:
$$
frac{|b-s|}t+frac b{s+t}+frac{b+t}sge 2 .
$$
Here, $b$ is allowed to vary between $0$ and $s$, and the first modulus is $|a|=-a=-(b-s)=s-b$, so we have to show:
$$
frac{s-b}t+frac b{s+t}+frac{b+t}sge 2 .
$$
But this is a linear (ergo convex) function in $b$, so the inequality has to be tested only for the possible extremities, $b=0$ and $b=s$. For $b=0$ we obtain $frac st+frac tsge 2$, true, with equality for $s=t$. For $b=s$ we get $frac s{s+t}+frac {s+t}sge 2$, an inequality of the same shape, but we cannot get equality. (For intermediate values of $b$ we get intermediate values of the expression.)
answered 10 hours ago
dan_fulea
5,8201312
add a comment |
up vote
0
down vote
up vote
0
down vote
I tried to give a simple proof by reducing to simpler and simpler cases, all the time doing "nothing", and the final rephrasing is simple, too.
First of all, let us note that the (LHS of the) given inequality is invariated by permutations of the letters / variables $a,b,c$. So we may and do assume $ale ble c$.
Let $s,t>0$ be then such that $b=a+s$, $c=b+t=a+s+t$.
Replacing the initial $a,b,c$ with $-c,-b,-a$, we may and do assume $bge 0$.
The given inequality can now be rewritten equivalently:
$$
S=
frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
ge 2 .
$$If $a$ is $ge0$, then all three numbers $a,b,c$ are $ge 0$, and we can write
$$
begin{aligned}
S
&= frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
\
&ge
frac{0}t +
frac{s}{s+t} +
frac{s+t}s
%ge 2sqrt{
%frac{s}{s+t} cdot
%frac{s+t}s}
ge2 ,qquad(age 0)
end{aligned}
$$
and the last $ge$ (AM-GM) is moreover $>0$ (because $tne 0$). The case with $a < b < c=a+s+tle 0$ is similar.So the single interesting case is the one with $a<0le b<c$ . We rewrite the inequality in the form:
$$
frac{|b-s|}t+frac b{s+t}+frac{b+t}sge 2 .
$$
Here, $b$ is allowed to vary between $0$ and $s$, and the first modulus is $|a|=-a=-(b-s)=s-b$, so we have to show:
$$
frac{s-b}t+frac b{s+t}+frac{b+t}sge 2 .
$$
But this is a linear (ergo convex) function in $b$, so the inequality has to be tested only for the possible extremities, $b=0$ and $b=s$. For $b=0$ we obtain $frac st+frac tsge 2$, true, with equality for $s=t$. For $b=s$ we get $frac s{s+t}+frac {s+t}sge 2$, an inequality of the same shape, but we cannot get equality. (For intermediate values of $b$ we get intermediate values of the expression.)
answered 10 hours ago
dan_fulea
5,8201312
I tried to give a simple proof by reducing to simpler and simpler cases, all the time doing "nothing", and the final rephrasing is simple, too.
First of all, let us note that the (LHS of the) given inequality is invariated by permutations of the letters / variables $a,b,c$. So we may and do assume $ale ble c$.
Let $s,t>0$ be then such that $b=a+s$, $c=b+t=a+s+t$.
Replacing the initial $a,b,c$ with $-c,-b,-a$, we may and do assume $bge 0$.
The given inequality can now be rewritten equivalently:
$$
S=
frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
ge 2 .
$$If $a$ is $ge0$, then all three numbers $a,b,c$ are $ge 0$, and we can write
$$
begin{aligned}
S
&= frac{|a|}t +
frac{|a+s|}{s+t} +
frac{|a+s+t|}s
\
&ge
frac{0}t +
frac{s}{s+t} +
frac{s+t}s
%ge 2sqrt{
%frac{s}{s+t} cdot
%frac{s+t}s}
ge2 ,qquad(age 0)
end{aligned}
$$
and the last $ge$ (AM-GM) is moreover $>0$ (because $tne 0$). The case with $a < b < c=a+s+tle 0$ is similar.So the single interesting case is the one with $a<0le b<c$ . We rewrite the inequality in the form:
$$
frac{|b-s|}t+frac b{s+t}+frac{b+t}sge 2 .
$$
Here, $b$ is allowed to vary between $0$ and $s$, and the first modulus is $|a|=-a=-(b-s)=s-b$, so we have to show:
$$
frac{s-b}t+frac b{s+t}+frac{b+t}sge 2 .
$$
But this is a linear (ergo convex) function in $b$, so the inequality has to be tested only for the possible extremities, $b=0$ and $b=s$. For $b=0$ we obtain $frac st+frac tsge 2$, true, with equality for $s=t$. For $b=s$ we get $frac s{s+t}+frac {s+t}sge 2$, an inequality of the same shape, but we cannot get equality. (For intermediate values of $b$ we get intermediate values of the expression.)
answered 10 hours ago
dan_fulea
5,8201312
answered 10 hours ago
dan_fulea
5,8201312
answered 10 hours ago
dan_fulea
5,8201312
answered 10 hours ago
dan_fulea
5,8201312
5,8201312
add a comment |
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Put everyting on one side, multiply everything by $|b-c||c-a||b-a|, and go from there.
– user3482749
yesterday