A proof to an old inequality

[ins-]

Let $a$, $b$, $c$ are positive reals such that $abc=1$. Prove that $\frac{a^2}{b^2+c}+\frac{b^2}{c^2+a}+\frac{c^2}{a^2+b} \ge \frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}$.

It is an old problem that was proven with computer using Buffalo way, after arqady observed it is applicable. Later I saw an attempt to be solved that I cannot understand (it is below). Can someone explain it?

\[\sum_{cyc}\frac{x^2}{y^2+z} \geq \frac{(x^2+y^2+z^2)^2}{\sum_{cyc} x^2y^2+x^2z} \geq\sum_{cyc} \frac{x}{y+z^2}\

[arqady]

Ваше неравенство очень тонкое и, скорее всего, предложенное решение (правое неравенство) неверно. Вечером скажу точно.
Даже если правое неравенство и верно, его всё равно надо как-то доказывать. То бишь, снова упираемся в BW.

[arqady]

Проверил. Правое неравенство верно. Красивое доказательство не вижу.

[ins-]

Thank you very much for the time and effort dedicated to this problem. Probably something different than BW can be discovered, but it is definitely not obvious at all. I suppose, it is hard to find even the chain of inequalities that I posted. It wasn't my discovery.

[sergei1961]

Интересный вопрос: из упомянутого в начале этой темы "старого" неравенства можно ухитриться вывести неравенство из этой темы:
topic106575.html

[ins-]

It is also very interesting and I started feeling myself like a blind, I cannot see it again. As I'm understanding you - it means the inequality I posted is stronger than the inequality from the link you posted. There is one problem - I'm not sure which one is harder to be proven.

About the left part of the chain I posted - it is easy to be proven. Nominator and denominator of the first fraction are multiplied by $x^2$. Similar operation is done for the other fractions. Then - Cauchy-Schwartz in Engel form. It remains the right part to be proven and the inequality is done.

[Rak so dna]

ins- как вы доказываете это неравенство с помощью BW ?

[ins-]

You can see the proof here: http://artofproblemsolving.com/communit ... 84p3514260 it is not obvious, even with using a computer.

[waxtep]

Not a complete solution at all, but might be of some interest: taking $a=b$ one gets
$LHS-RHS=\frac {(a-1)^2} {a^5(a^5+1)(a^4+1)} P_{13}^+$,
where $P_{13}^+$ is a $13^{th}$ degree polynomial with all positive coefficients; it can be found w/o computer, as it is just $a^{14}+a^{13}+a^{10}-a^9+a^8-a^5-a^4-1$ divided by its easily seen divisor $(a-1)$. Now the hard part is to show that $a=b$ is the only extremum, not sure is there any natural way to complete this...

[waxtep]

...similarly, $b=\frac 1 a$ yields $LHS-RHS=\frac {(a-1)^2} {\ldots^+} P_7^+$, but how to get there... I hope to find some $b=b(a,k)$ having a clear extremum in terms of $k$, but to no success yet.

[ins-]

The way you deal with the problem is interesting, but the proofs when the minimum is achieved seem to not be easy.