YadisBeles wrote: Let $a,b,c$ be nonnegative real numbers such that $(a+b)(b+c)(c+a) \neq0$. Find the minimum of $(a+b+c)^{2016}(\frac{1}{a^{2016}+b^{2016}}+\frac{1}{b^{2016}+c^{2016}}+\frac{1}{c^{2016}+a^{2016}})$. Source Let $a,b,c$ be nonnegative real numbers such that $(a+b)(b+c)(c+a) \neq0$.Then $$(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\ge \frac{9}{2}.$$

Let $a,b,c$ be nonnegative real numbers such that $(a+b)(b+c)(c+a) \neq0$.Prove that $$(a+b+c)^2(\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2})\ge 10$$$$(a+b+c)^3(\frac{1}{a^3+b^3}+\frac{1}{b^3+c^3}+\frac{1}{c^3+a^3})\ge 20$$

sqing wrote: Let $a,b,c$ be nonnegative real numbers such that $(a+b)(b+c)(c+a) \neq0$.Prove that $$(a+b+c)^2(\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2})\ge 10$$$$(a+b+c)^3(\frac{1}{a^3+b^3}+\frac{1}{b^3+c^3}+\frac{1}{c^3+a^3})\ge 20$$ Generalization Let $a,b,c$ be nonnegative real numbers such that $(a+b)(b+c)(c+a) \neq0$,$\alpha \ge 2$.Prove that $$\left( {x + y + z} \right)^\alpha \left( {\frac{1}{{x^\alpha + y^\alpha }} + \frac{1}{{y^\alpha + z^\alpha }} + \frac{1}{{z^\alpha + x^\alpha }}} \right) \ge 5 \times 2^{\alpha - 1} {\rm{ }}$$ see http://blog.sina.com.cn/s/blog_547541ce0102wena.html

sqing wrote: Let $a,b,c$ be nonnegative real numbers such that $(a+b)(b+c)(c+a) \neq0$.Prove that

$$(a+b+c)^2(\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2})\ge 10$$**

http://www.artofproblemsolving.com/community/c6h1122010p5166529 http://www.artofproblemsolving.com/community/c6h1207618p5970656 Let $a,b,c \ge 0$ such that $ab+bc+ca>0$ and $0\le k\le 11.$ Prove that \[ \frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\frac{k}{a^2+b^2+c^2} \ge \frac{10+2k}{(a+b+c)^2}\]

Because the inequality is homogeneous, WLOG: $a+b+c=2$, We now prove for all $k\geq 2$ That $f(a,b,c)=\frac{1}{a^{k}+b^{k}}+\frac{1}{b^{k}+c^{k}}+\frac{1}{c^{k}+a^{k}}\geq \frac{5}{2}$

First, little lemma: if $k\geq 2$ and $0 \leq x\leq 1$ then $(1+\frac{x}{2})^k\geq 1+ x^k$proof

Now WLOG $min\{a,b,c\}=c$, we prove that $f(a,b,c)\geq f(a+\frac{c}{2},b+\frac{c}{2},0)$: It is clear that $\frac{1}{a^k +b^k} \geq \frac{1}{(a+ \frac{c}{2})^k+(b+ \frac{c}{2})^k}$ On the other hand set $\frac{c}{a}=x\leq 1$ so $\frac{1}{a^k +c^k}\geq \frac{1}{(a+ \frac{c}{2})^k} \Leftrightarrow (a+ \frac{c}{2})^k\geq a^k +c^k \Leftrightarrow (1+ \frac{x}{2})^k\geq 1+x^k$ which is lemma 1. The same thing can be similarly said for $b,c$. So knowing this, it is enough to prove for $c=0$: We have that $a+b=2$ so by AM-GM $ab\leq 1$ now: $$\frac{1}{a^k+b^k} + \frac{1}{a^k} + \frac{1}{b^k}= \frac{1}{a^k+b^k} + \frac{a^k+b^k}{(ab)^k} = (\frac{1}{a^k+b^k} + \frac{a^k+b^k}{4(ab)^k})+ \frac{3(a^k+b^k)}{4(ab)^k} \geq \frac{1}{(ab)^{\frac{k}{2}}} +\frac{3}{2(ab)^{\frac{k}{2}}} \geq \frac{5}{2}$$Where the last inequality is because $k\geq 2$ and $ab\leq 1$. The equality case is one of them $0$ and the other two equal.