I have a hard problem this is it :
Let $a,b,c>0$ such that $a^ab^bc^c=1$ then we have :
$$\Big(\frac{1}{a^2+b^2}\Big)^2+\Big(\frac{1}{b^2+c^2}\Big)^2+\Big(\frac{1}{c^2+a^2}\Big)^2\geq \frac{3}{4}$$
I try to use Jensen's ienquality applied to the function $f(x)=\frac{1}{x^2}$ but it doesn't works for all the values .
if we apply Am-Gm it's like Jensen's inequality so I forget this ways .
Maybe If we apply Karamata's inequality but I didn't found the right majorization .
I try also to use Muirhead inequality but without success .
So I'm a bit lost with that if you have a hint or a full answer I will be happy to read your work .
Best Answer
Nelson Faustino guessed that $(a^2+b^2)(b^2+c^2)(c^2+a^2)\le 8$. Let us prove it.
Proof: We will use the following lemma whose proof is given later.
Lemma 1: Let $a, b, c\ge 0$ with $11(a^2+b^2+c^2) + 4(a+b+c) - 45 \le 0$. Then $(a^2+b^2)(b^2+c^2)(c^2+a^2)\le 8$.
Let us begin. Using AM-GM, we have $$\frac{1}{(a^2+b^2)^2} + \frac{1}{(b^2+c^2)^2} + \frac{1}{(c^2+a^2)^2} \ge 3\sqrt[3]{\frac{1}{(a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2}}.$$ It suffices to prove that $(a^2+b^2)(b^2+c^2)(c^2+a^2)\le 8$. To this end, we have \begin{align} &a, b, c > 0, \quad a^ab^bc^c=1 \qquad\qquad (1)\\ \Longrightarrow \quad & a\ln a + b\ln b + c\ln c = 0, \quad 0 < a, b, c < \frac{159}{100}\qquad\qquad (2)\\ \Longrightarrow \quad &11(a^2+b^2+c^2) + 4(a+b+c) - 45 \le 0, \quad a, b, c > 0. \qquad\qquad (3) \end{align} Here, in (2), since $-u\ln u \le \frac{1}{\mathrm{e}}$ for $u>0$, we have $a\ln a, b\ln b, c\ln c \le \frac{2}{\mathrm{e}}$ which, when combined with $\frac{159}{100}\ln \frac{159}{100} > \frac{2}{\mathrm{e}}$, results in $a, b, c < \frac{159}{100}$ (noting that $x \mapsto x\ln x$ is strictly increasing for $x > 1$); In (3), we have used the fact that $$x\ln(x) \ge \frac{11}{26}x^2 + \frac{2}{13}x - \frac{15}{26}, \quad \forall 0 < x < \frac{159}{100}.\qquad (4)$$ Then, according to Lemma 1, the desired result follows. We are done.
Proof of (4): It suffices to prove that $$\ln x \ge \frac{\frac{11}{26}x^2 + \frac{2}{13}x - \frac{15}{26}}{x}, \quad \forall 0 < x < \frac{159}{100}.$$ Let $h(x) = \ln x - \frac{\frac{11}{26}x^2 + \frac{2}{13}x - \frac{15}{26}}{x}$. We have $h'(x) = -\frac{11(x-1)(x-\frac{15}{11})}{26x^2}$. Thus, $h(x)$ is decreasing on $(0, 1)$, increasing on $(1, \frac{15}{11})$ and decreasing on $(\frac{15}{11}, \infty)$. Note that $h(1) = 0$ and $h(\frac{159}{100}) > 0 $. The desired result follows.
Proof of Lemma 1: One of the methods is the uvw method as follows. However, I hope to see nice proofs of Lemma 1.
Let $a+b+c = 3u$, $ab+bc+ca = 3v^2$ and $abc = w^3$.
The constraint $11(a^2+b^2+c^2) + 4(a+b+c) - 45 \le 0$ becomes $11(9u^2 - 6v^2) + 12u - 45 \le 0$. The constraint does not involve $w^3$.
We need to prove that $f(w^3) = 8 - (a^2+b^2)(b^2+c^2)(c^2+a^2) = w^6+6u(9u^2-6v^2)w^3-81u^2v^4+54v^6+8 \ge 0$. Since $9u^2 - 6v^2 \ge 0$ and $w^3\ge 0$, $f(w^3)$ is increasing with $w^3$. Thus, we only need to prove the case when $a=b$ or $c=0$. The rest is not hard and thus omitted.