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!
With computer, here is an ugly solution:

WLOG, assume that $c = \min(a, b, c).$ 
Since $x^{2/3}, x \ge 0$ is concave, we have
\begin{align}
\frac{b}{\sqrt[3]{b+c}} + \frac{c}{\sqrt[3]{c+a}}
&amp;= \frac{b}{b+c}(b+c)^{2/3} + \frac{c}{c+a}(c+a)^{2/3}\\
&amp;\le \Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^{2/3}.
\end{align}
On the other hand, we have
$$\Big(\frac{1+5a+5b}{5+a+b}\Big)^3 - (a+b)^2 = \frac{(a^2+2ab+b^2+18a+18b+1)(1-a-b)^3}{(5+a+b)^3} \ge 0$$
which results in
$(a+b)^{2/3} \le \frac{1+5a+5b}{5+a+b}$.

Thus, it suffices to prove that
$$\frac{a}{a+b}\frac{1+5a+5b}{5+a+b} +  \Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^{2/3} \le \frac{31}{27}.$$
It suffices to prove that
$$\Big(\frac{31}{27} - \frac{a}{a+b}\frac{1+5a+5b}{5+a+b}\Big)^3\ge
\Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)^3\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^2. $$
After homogenization, it suffices to prove that
$$(a+b+c)^2\Big(\frac{31}{27} - \frac{a}{a+b}\frac{(a+b+c)+5a+5b}{5(a+b+c)+a+b}\Big)^3\ge
\Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)^3\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^2.$$
If suffice to prove that $f(a,b,c) \ge 0$ for $a, b, c \ge 0; \ c = \min(a,b,c)$ where $f(a,b,c)$ is a polynomial (a long expression).

We apply the Buffalo Way. There are three possible cases:

1) If $c = 0$, we have $f(a,b, 0) = 216a(a+28b)(8a-19b)^2(a+b)^5 \ge 0$.

2) If $c &gt; 0$ and $c \le a \le b$, let $a = c + s, \ b = c+s+t; \ s,t \ge 0$.
$f(c+s, c+s+t, c)$ is a polynomial in $c, s, t$ with non-negative coefficients. 

3) If $c&gt;0$ and $c \le b\le a$, let $b = c+s, \ a = c+s + t; \ s,t \ge 0$.
We have 
\begin{align}
&amp;f(c+s+t, c+s, c) = g(c,s,t) + 216(s+t)(29s+t)(11s-8t)^2(2s+t)^5\\
&amp;\quad +  108(146777s^4+13368s^3t-97257s^2t^2+42286st^3+7200t^4)(2s+t)^4c
\end{align}
where $g(c,s,t)$ is a polynomial with non-negative coefficients. Note that
$$146777s^4+13368s^3t-97257s^2t^2+42286st^3+7200t^4 \ge 2\sqrt{146777 \cdot 7200} s^2 t^2 + 2\sqrt{13368\cdot 42286}s^2t^2 - 97257s^2t^2 \ge 0.$$

The desired result follows. We are done.
I have an half proof :
We have :

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

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