<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] By Holder $$\left(\sum_{cyc}\frac{a^3}{(a+b)^3}\right)^2\sum_{cyc}1\geq\left(\sum_{cyc}\sqrt[3]{\left(\frac{a^3}{(a+b)^3}\right)^2\cdot1}\right)^3=\left(\sum_{cyc}\frac{a^2}{(a+b)^2}\right)^3.$$
Thus, it's enough to prove that
$$\frac{\left(\sum\limits_{cyc}\frac{a^2}{(a+b)^2}\right)^3}{3}\geq\frac{9}{64}$$ or
$$\sum\limits_{cyc}\frac{a^2}{(a+b)^2}\geq\frac{3}{4}.$$
Now, by C-S
$$\sum\limits_{cyc}\frac{a^2}{(a+b)^2}=\sum\limits_{cyc}\frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a+b)^2(a+c)^2}.$$
Thus, it's enough to prove that
$$4\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a+b)^2(a+c)^2,$$
which is true even for all reals $a$, $b$ and $c$.
Indeed, the last inequality is symmetric inequality by degree four, 
which says that by $uvw$ (https://math.stackexchange.com/tags/uvw/info )
it's enough to prove the last inequality for equality case of two variables and since 
it's the homogeneous inequality by even degree, we can assume $b=c=1$, which gives
$$(a-1)^2(a+3)^2\geq0.$$
Done!
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