Non - symetric inequality in 3 variables

88 Views Asked by At

How can we prove the following inequality for $a,b,c>0$:

$$\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}\leq \dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{c^2+a^2} \ ?$$

2

There are 2 best solutions below

0
On BEST ANSWER

A full expending gives $$\sum\limits_{cyc}(2a^4b^2+a^4c^2-a^4bc-a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0$$ Since by Muirhead $\sum\limits_{cyc}(a^4b^2+a^4c^2-a^4bc-a^3b^3)\geq0$ is true,

it remains to prove that $\sum\limits_{cyc}(a^4b^2-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0$ or $$\sum\limits_{cyc}\left(\frac{a^2}{c^2}-\frac{a}{c}-\frac{c}{a}+1\right)\geq0$$

Now, let $\frac{a}{c}=x$, $\frac{b}{a}=y$ and $\frac{c}{b}=z$.

Hence, it remains to prove that $x^2+y^2+z^2-x-y-z-xy-xz-yz+3\geq0$, where $xyz=1$.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $9u^2-9v^2-3u+3\geq0$ or, $3u^2-3v^2-uw+w^2\geq0$,

which is a liner inequality of $v^2$,

which says that it's enough to prove the last inequality for an extremal value of $v^2$.

But $x$, $y$ and $z$ are positive roots of the equation $(X-x)(X-y)(X-z)=0$ or $$X^3-3ux^2+3v^2x-w^3=0$$ or $$3v^2x=-x^3+3ux^2+w^3$$

which says that a line $y=3v^2x$ and a graph $y=-x^3+3ux^2+w^3$ have three common points and $v^2$ gets an extremal value for an equality case of two variables (draw it!).

Let $y=x$ and $z=\frac{1}{x^2}$.

We need to prove that $(x-1)^2(x^4+2x^2+2x+1)\geq0$.

Done!

0
On

Also we can use the following way. We need to prove that $$\sum\limits_{cyc}\left(\frac{2a^2}{a^2+b^2}-1\right)+\sum\limits_{cyc}\left(1-\frac{2ab}{a^2+b^2}\right)\geq0$$ or $$\sum\limits_{cyc}\frac{(a-b)^2}{a^2+b^2}\geq\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{\prod\limits_{cyc}(a^2+b^2)}$$

By C-S $\sum\limits_{cyc}\frac{(a-b)^2}{a^2+b^2}\geq3\sqrt[3]{\prod\limits_{cyc}\frac{(a-b)^2}{a^2+b^2}}$ and it remains to prove that $$27\prod\limits_{cyc}(a^2+b^2)^2\geq(a-b)(b-c)(c-a)\prod\limits_{cyc}(a+b)^3$$ and since $a^2+b^2\geq\frac{1}{2}(a+b)^2$, it remains to prove that $$27\prod\limits_{cyc}(a^2+b^2)\geq8\prod\limits_{cyc}(a^2-b^2)$$ which is obviously true.

Done!