<s>[INST] Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$ When $a,b,c > 0$, prove
$$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c ^{\frac34}}{2^{\frac34}}$$
I tried the substitution $x=a^4,\ldots$ but I have no idea how to deal with the left- hand side. I tried some C-S but it goes nowhere. I think Bernoulli's inequality may be the only way to prove this inequality.
 [/INST] Version of 29.06.18
HINT

$$\mathbf{\color{brown}{Task\ transformations}}$$
Let
$$
\begin{cases}
&b^3+c^3= 2x^4\\
&c^3+a^3= 2y^4\\
&a^3+b^3= 2z^4
\end{cases}\Rightarrow
\begin{cases}
a^3=-x^4+y^4+z^4\\
b^3= x^4-y^4+z^4\\
c^3= x^4+y^4-z^4\tag1,
\end{cases}$$
then the issue equation transforms to
$$\rlap\bigcirc\!\sum\dfrac{-x^4+y^4+z^4}{z^3}\ge \rlap\bigcirc\!\sum\sqrt[4]{-x^4+y^4+z^4}.$$
$$\rlap\bigcirc\!\sum z\dfrac{-x^4+y^4+z^4}{z^4}\left(1-\dfrac1{\left(1+\dfrac{y^4-x^4}{z^4}\right)^{3/4}}\right)\ge 0.\tag2$$
Using inequality
$$(1+t)^\alpha\le 1+\alpha t,\quad 1>\alpha>0,\quad t\ge-1,\tag3$$
for $\alpha=\dfrac34,$ can be obtained the stronger inequality than $(2):$
$$\rlap\bigcirc\!\sum z\left(1+\dfrac{y^4-x^4}{z^4}\right)\left(1-\dfrac1{1+\dfrac34\dfrac{y^4-x^4}{z^4}}\right)\ge 0,$$
$$\rlap\bigcirc\!\sum \dfrac{y^4-x^4}{z^3}\dfrac{z^4+y^4-x^4}{4z^4+3y^4-3x^4}\ge 0.\tag4$$

$$\mathbf{\color{brown}{Conditions}}$$
Taking in account $(1),$ the boundary conditions are
$$0<x<\sqrt[4]{y^4+z^4},\quad 0<y<\sqrt[4]{z^4+x^4},\quad 0<z<\sqrt[4]{x^4+y^4}.\tag5$$
Taking in account $(5)$, term of $LSH(4)$ is negative iff $x > y.$ If $0<x\le y\le z,$ then the inequality $(4)$ is satisfied.
The task $(4)-(5)$ has rotational symmetry, so WLOG it is required to check only the case
$$x\ge y\ge z>0.\tag6$$
The obtained task $(4)-(6)\ $ is correct and allows a simple proof.
</s>