<s>[INST] Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$ 
Let $a,b,c$ be positive real numbers. Prove that $$\Big({a\over a+b}\Big)^3+\Big({b\over b+c}\Big)^3+ \Big({c\over c+a}\Big)^3\geq {3\over 8}$$

If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and
$$\Big({1\over 1+x}\Big)^3+\Big({1\over 1+y}\Big)^3+ \Big({1\over 1+z}\Big)^3\geq {3\over 8}$$
Since $f(x)=\Big({1\over 1+x}\Big)^3$ is convex we get, by Jensen,: $$\Big({1\over 1+x}\Big)^3+\Big({1\over 1+y}\Big)^3+ \Big({1\over 1+z}\Big)^3\geq 3f({x+y+z\over 3})$$
Unfortunately, since $f$ is decreasing we don't have $f({x+y+z\over 3}) \geq f(1) = {1\over 8}$.
Some idea how to solve this?
 [/INST] This is more of a comment,  but I don't have the reputation.  Use Lagrange multipliers.  Solving, you find that the critical points occur when $xyz=1$ and $yz (1+x)^4 = xz (1+y)^4 = xy (1+z)^4$.  I think the only solution is $x=y=z=1$.  Clearly, it's a minimum and plugging in shows the bound.  
Added:
We can rewrite the condition as $xyz=1$ and $\frac{(1+x)^4}{x} = \frac{(1+y)^4}{y} = \frac{(1+z)^4}{z}$
The function $g(x)=\frac{(1+x)^4}{x}$ is decreasing from $0$ to $1/3$ and increasing from $1/3$ to $\infty$.  This shows that 2 of $x,y,z$ must be equal (WLOG $x$ and $y$) and $z=1/x^2$.  It remains to solve $\frac{(1+x)^4}{x} = \frac{(1+1/x^2)^4}{1/x^2}$.  This time, it's not hard to check $x=1$ is the only solution and we are done.
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