If $a+b+c=1$ then $\sum\limits_{cyc}\frac{a}{\sqrt[3]{a+b}}\leq\frac{31}{27}$ 
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$ and $a+b+c=1$.
Prove that:
$$\frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}\leq\frac{31}{27}$$

The equality occurs for $(a,b,c)=\left(\frac{19}{27},\frac{8}{27},0\right)$.
This inequality is similar to the following inequality, which was proposed by Walther Janous.

For all non-negatives $x$, $y$ and $z$ such that $xy+xz+yz\neq0$ prove that:
$$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq\frac{5}{4}\sqrt{x+y+z}$$

My proof:
By Cauchy-Schwarz $$\left(\sum_{cyc}\frac {x}{\sqrt {x+y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}.$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$$ or
$$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$$
$$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$$ or
$$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$$
$$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0,$$ which is obvious.
If we want to use a similar way for the starting inequality, we need to use Holder, which gives very big numbers.
Maybe there is another reasoning?
Thank you!
 A: Here is my personal take on the problem;
We can use the substitutions $b+c = 1-a$, $c+a = 1-b$, $c = 1-a-b$ to reduce the problem to the following inequality in arbitrary $a,b$;
$$\dfrac{a}{\sqrt[3]{a+b}} + \dfrac{b}{\sqrt[3]{1-a}} + \dfrac{1-a-b}{\sqrt[3]{1-b}} \leq \dfrac{31}{27}$$
Define $f(a,b) = \dfrac{a}{\sqrt[3]{a+b}} + \dfrac{b}{\sqrt[3]{1-a}} + \dfrac{1-a-b}{\sqrt[3]{1-b}}$. I suspect that using the first and second derivative tests to find local maxima-minima could solve the problem from here, as $\dfrac{\partial f}{\partial a}$ and $\dfrac{\partial f}{\partial b}$ are, while messy, not difficult to calculate.
In fact, this can be generalized quite a bit. This method can be generalized as follows;
Consider $\sum_{cyc} g(x_1,\dots,x_n)$ for some well-defined function $g$, where the cyclic behavior occurs over a set of variables $x_1,\dots,x_k$ for $k > n$ and we restrict these $x$ by stating that $\sum_{j=1}^k x_j = A$ for a fixed constant $A$. Then we can write $x_k = A - \sum_{j=1}^{k-1} x_j$ and use this as the basis for a substitution to create a function $f(x_1,\dots,x_{k-1}) = \sum_{cyc} g(x_1,\dots,x_n)$ with restrictions on the $x_j$ removed and using the substitution previously given for $x_k$. If $g$ is differentiable, then so will $f$ be, and so local extremes can be calculated. If $g$ is twice differentiable, then so is $f$ and we can use the second derivative test to test whether these values are true extrema or saddle points.
A: I have an half proof :
We have :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have :
$$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}}$$

We need a lemma :

Lemma
Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then  we have :
$$\sum_{cyc}0.5(a+b)^{\frac{2}{3}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}}$$

Proof : It's Jensen's inequality apply to the concave function $x^{\frac{2}{3}}$.
Now we prove :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then  we have :
$$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \frac{a+b}{2(a+b)^{\frac{1}{3}}}+\frac{b+c}{2(b+c)^{\frac{1}{3}}}+\frac{a+c}{2(a+c)^{\frac{1}{3}}}$$

Proof : The inequality is equivalent to :
$$\frac{b-a}{2(a+b)^{\frac{1}{3}}}+\frac{c-b}{2(c+b)^{\frac{1}{3}}}+\frac{a-c}{2(a+c)^{\frac{1}{3}}}\leq 0$$
Wich is true with SOS method or rearrangement inequality .
Finally we have :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then  we have :
$$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}} <\frac{31}{27}$$

Conclusion :
So the real case to threat is when $b\geq a\geq c $.
