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!

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.
