gavrilos wrote: Let $a,b,c$ be positive real numbers satisfying $a+b+c=3$.Prove that $\sum_{sym} \frac{a^{2}}{(b+c)^{3}}\geq \frac{3}{8}$ By C-S $\sum_{cyc}\frac{a^{2}}{(b+c)^{3}}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a^2(b+c)^3}$. Hence, it remains to prove that $8(a^2+b^2+c^2)^2(a+b+c)\geq9\sum\limits_{cyc}a^2(b+c)^3$, which is fifth degree. Hence, by $uvw$ we need to check two cases: 1. $b=1$ and $c\rightarrow0^+$, which gives something obvious; 2. $b=c=1$, which gives $(a-1)^2(4a^3+16a^2+35a+23)\geq0$.

It is simpler arqady By C-S and Nesbitt we have that \[2(a+b+c)\sum_{cyc}\frac{a^2}{(b+c)^3}\geq\left(\sum_{cyc}\frac{a}{b+c}\right)^2\geq\frac{9}{4}\]

Let $f(x) = \frac{x^2}{(3-x)^3}$. f is convex over [0,3], then by Jensen, we have: \[ f(a) + f(b) +f(c) \geq 3f(\frac{a+b+c}{3})=\frac{3}{8}\]

Nice your proofs, silouan and bel.jad5! See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3567758#p3567758

Jenson inequality

Tangent line also works here. $\sum\frac{a^2}{(3-a)^3}\ge\sum\left(\frac{7a}{16}-\frac{5}{16}\right)=\frac{3}{8}$

Let $a,b,c$ be positive real numbers. Prove that $$\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}\geq \frac{9}{8(a+b+c)}$$

sqing wrote: Let $a,b,c$ be positive real numbers. Prove that $$\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}\geq \frac{9}{8(a+b+c)}$$ My proof is proving your question Mr. Sqing.

Nice problem..... making the following change (I think it was unnecessary, in the end. ) $a=3k, b=3m, c=3n$, so $k+m+n=1$ the problems is equivalent to: \[\sum_{sym} \frac{(3k)^2}{3^3(m+n)^3}=\sum_{sym}\geq \frac{3}{8}\] \[\sum_{sym} \frac{k^2}{(1-k)^3}\geq \frac{9}{8}\] Let the function $f=\dfrac{x^2}{(1-x)^3}$, is convex in the interval $[0,1]$, so by jensen's inequality \[\sum_{sym} \frac{k^2}{(1-k)^3} = f(k)+f(m)+f(n)\geq 3 f(\dfrac{k+m+n}{3})=3f(\dfrac{1}{3})=\dfrac{9}{8}\] As desired.

Version 1- 4 Variable Let $a,b,c$ be positive reels. Then prove that $$\dfrac{a^2}{(b+c)^3}+\dfrac{b^2}{(c+d)^3}+\dfrac{c^2}{(d+a)^3}+\dfrac{d^2}{(a+b)^3}\geq \dfrac{2}{a+b+c+d}$$

Generalization 2 Let $a,b,c$ be positive reels ($k\geq 1$) such that $n\geq 0$ is reel. Then prove that $$\dfrac{a^k}{(b+c)^{n+k}}+\dfrac{b^k}{(c+a)^{n+k}}+\dfrac{c^k}{(a+b)^{n+k}}\geq \dfrac{3^{n+1}}{2^{n+k}(a+b+c)^n}$$

This is also a generalization of Nesbitt's original inequality when you give $k=1,n=0$. See Nesbitt- Power Generalization Nesbitt- Power Generalization

Generalization 3 Let $a,b,c,d$ ve positive reels ($k\geq 1$) such that $n\geq 0$ is reel. Then prove that $$\dfrac{a^k}{(b+c)^{n+k}}+\dfrac{b^k}{(c+d)^{n+k}}+\dfrac{c^k}{(d+a)^{n+k}}+\dfrac{d^k}{(a+b)^{n+k}}\geq \dfrac{2^{n+k-2}}{(a+b+c+d)^n}$$