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.
 A: Let $x_1=a^3$, $x_2=b^3$, $x_3=c^3$, and $S=\sum_i x_i$. 
We wish to prove
$$ \frac{x_1}{(S-x_3)^{3/4}}+\frac{x_2}{{(S-x_1)}^{3/4}}+\frac{x_3}{{(S-x_2)}^{3/4}}-\frac{\sum_i x_i^{1/4}}{2^{3/4}}\ge 0.$$
We consider the problem to minimize $ \frac{x_1}{(S-x_3)^{3/4}}+\frac{x_2}{{(S-x_1)}^{3/4}}+\frac{x_3}{{(S-x_2)}^{3/4}}-\frac{\sum_i x_i^{1/4}}{2^{3/4}}$ over variables $x_1,x_2, x_3$, such that $x_i\ge 0$ and $\sum_i x_i = S$. We see that all $x_i$ equal satisfies KKT and is indeed the minimal choice (If Largrange multiplier for any of $x_i>0$ is non-zero, $x_i=0$ and the condition holds and thus we can assume that the Lagrange for each $x_i$ is zero and thus there is a single Lagrange multiplier for the sum and equal $x_i$ trivially satisfies the conditions.). Thus, we get the above expression to hold.
A: Define:
$$
x = \frac{a^{3/4}}{2^{3/4}} = \left(\frac{a}{2}\right)^{3/4} \quad ; \quad
y = \frac{b^{3/4}}{2^{3/4}} = \left(\frac{b}{2}\right)^{3/4} \quad ; \quad
z = \frac{c^{3/4}}{2^{3/4}} = \left(\frac{c}{2}\right)^{3/4}
$$
Then:
$$
a = 2\,x^{4/3} \quad ; \quad b = 2\,y^{4/3} \quad ; \quad c = 2\,z^{4/3}
$$
And:
$$
\left(a^4\right)^{3/4} = a^3 = 2^3 x^4 \quad ; \quad
\left(b^4\right)^{3/4} = b^3 = 2^3 y^4 \quad ; \quad 
\left(c^4\right)^{3/4} = c^3 = 2^3 z^4
$$
Finally, when $x,y,z > 0$, prove:
$$
f(x,y,z) = 
\frac{2^3 x^4}{\left(2^3 x^4 + 2^3 y^4\right)^{3/4}} +
\frac{2^3 y^4}{\left(2^3 y^4 + 2^3 z^4\right)^{3/4}} +
\frac{2^3 z^4}{\left(2^3 z^4 + 2^3 x^4\right)^{3/4}} \ge 1
$$
Where it can be assumed without loss of generality that: $\,x+y+z = 1$ .
The problem is thus reduced to a familiar one, quite similar to:

$x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
Olympiad Inequality $\sum_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$

And can be treated accordingly:


The minimum of our function inside the abovementioned triangle must shown to be greater or equal to one.
Another proof without words is attempted
by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as:

nivo := min + g/grens*(max-min); { grens = 20 ; g = 1..grens }

The whiteness of the isolines is proportional to the (positive) function values; they are almost black near the minimum and almost white near the maximum values.
Maximum and minimum values of the function are observed to be:


 1.00001285611974E+0000 < f < 1.68177794816992E+0000

The little $\color{blue}{\mbox{blue}}$ spot in the middle is where $\,\left| f(x,y,z) - 1 \right| < 0.001$ . Due to symmetry,
an absolute minimum of the function is expected indeed at $(x,y,z) = (1/3,1/3,1/3)$.

Note.
Conditions similar to $\;x+y+z=1\;$ often occur in these inequalities, whether that is explicitly or implicitly.
An explicit example has been provided with another
HN_NH question :
Prove $\frac{xy}{5y^3+4}+\frac{yz}{5z^3+4}+\frac{zx}{5x^3+4} \leqslant \frac13$
The current inequality is an example of implicit occurrence. Let the function $f(x,y,z)$
be defined as above, then we have the equivalent inequality $\;f(x,y,z) \geqslant x+y+z$ .
It is clear that $f$ has the following property for all real $\lambda > 0$ :
$\;f(\lambda x,\lambda y,\lambda z) = \lambda f(x,y,z) \geqslant \lambda (x+y+z)$ .
Therefore $\lambda$ has no influence whatsoever on the inequality being true or false; we can always divide $x,y,z$ by a factor $\lambda$ such that $x+y+z=1$ .
Thus enabling a triangle mapping method once again.
A: It's not a proof, but I think it can help.
We can rewrite this inequality in the following form.
$$\sum_{cyc}\frac{a^4}{\sqrt[4]{\left(\frac{a^4+b^4}{2}\right)^3}}\geq a+b+c,$$
where $a$, $b$ and $c$ are positives.
Now, since for $(a,b,c)=(1.98,0.89,1.38)$
$$\sum_{cyc}\frac{a^4}{\sqrt[4]{\left(\frac{a^4+b^4}{2}\right)^3}}-a-b-c=0.0080...,$$
we can use the  Holder's inequality.
I checked that the following Holder does not help.
$$\left(\sum_{cyc}\frac{a^4}{\sqrt[4]{(a^4+b^4)^3}}\right)^4\sum_{cyc}a^4(a^4+b^4)^3(a+mb+nc)^5\geq$$
$$\geq\left(\sum_{cyc}a^4(a+mb+nc)\right)^5$$
because the inequality
$$8\left(\sum_{cyc}a^4(a+mb+nc)\right)^5\geq(a+b+c)^4\sum_{cyc}a^4(a^4+b^4)^3(a+mb+nc)^5$$ is wrong for all $m\geq0$ and $n\geq0$.
By the way, I think the following Holder can help.
$$\left(\sum_{cyc}\frac{a^4}{\sqrt[4]{(a^4+b^4)^3}}\right)^4\sum_{cyc}a^4(a^4+b^4)^3(a^2+kb^2+mc^2+lab+nac+pbc)^5\geq$$
$$\geq\left(\sum_{cyc}a^4(a^2+kb^2+mc^2+lab+nac+pbc)\right)^5,$$
where $a^2+kb^2+mc^2+lab+nac+pbc>0$ for all positives $a$, $b$ and $c$.
It's enough to prove that
$$8\left(\sum_{cyc}a^4(a^2+kb^2+mc^2+lab+nac+pbc)\right)^5\geq$$
$$\geq(a+b+c)^4\sum_{cyc}a^4(a^4+b^4)^3(a^2+kb^2+mc^2+lab+nac+pbc)^5.$$
Now, we can substitute in the last inequality $(a,b,c)=(1.98,0.89,1.38)$
and we can choose parameteres  $k,$ $l$, $m$, $n$ and $p$  such that the inequality is true.
I hope that it's possible! I think it's possible even for one or more of these parameters equal to zero. For this we need some software, which I have no. 
After choosing of parameters we can try to prove the inequality by BW, for which we need software again.
About BW see here: https://artofproblemsolving.com/community/c6h522084
After proving of the starting inequality we can say that this inequality we can prove by hand, but after some days of idiotic computations. 
A: 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.
A: Partial answer :
Hint :
Using WRCF theorem see https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/1029-242X-2011-101
We have $c=\min{(a,b,c)}$ and let $f(x)$ such that :
$$f\left(x\right)=\left(\frac{1}{1+x^{3}}\right)^{\frac{3}{4}}$$
It seems we have $f''(x)>0$ for $x\geq 1$
For  and $a,b,c\in[1,2]$ it seems we have :
$$g\left(x\right)=\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}f\left(x\right)+\left(1-\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}\right)f\left(\frac{\left(1-\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}x\right)}{1-\frac{c^{\frac{3}{4}}}{a^{\frac{3}{4}}+b^{\frac{3}{4}}+c^{\frac{3}{4}}}}\right)-f\left(1\right)\geq 0$$
Where $0.5\leq x\leq 1$ .
