TITLE: Prove that $\sqrt[3]{\frac{a}{a+7b}}+\sqrt[3]{\frac{b}{b+7a}}\geq\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}}$
QUESTION [6 upvotes]: Let $a>0$ and $b>0$. Prove that:
  $$\sqrt[3]{\frac{a}{a+7b}}+\sqrt[3]{\frac{b}{b+7a}}\geq\sqrt{\frac{a}{a+3b}}+\sqrt{\frac{b}{b+3a}}$$

This inequality was on one of old Chinese contests.
If $a=xb$ so we get an inequality of one variable, but it's not comforting. 
I tried also the following.
Let $a^2+b^2=2xab$. Hence, $x\geq1$ and we need to prove that
$$\sqrt[3]{\frac{a}{a+7b}}+\sqrt[3]{\frac{b}{b+7a}}-\sqrt{\frac{a}{a+3b}+\frac{b}{b+3a}+2\sqrt{\frac{ab}{(a+3b)(b+3a)}}}\geq0$$ or
$$\sqrt[3]{\frac{a}{a+7b}}+\sqrt[3]{\frac{b}{b+7a}}-\sqrt{\frac{3a^2+3b^2+2ab}{3a^2+3b^2+10ab}+2\sqrt{\frac{ab}{3a^2+3b^2+10ab}}}\geq0$$ or
$$\sqrt[3]{\frac{a}{a+7b}}+\sqrt[3]{\frac{b}{b+7a}}-\sqrt{\frac{3x+1}{3x+5}+\sqrt{\frac{2}{3x+5}}}\geq0$$ or
$$\frac{7x+1}{7x+25}-\sqrt{\left(\frac{3x+1}{3x+5}+\sqrt{\frac{2}{3x+5}}\right)^3}+\frac{3\sqrt{\frac{3x+1}{3x+5}+\sqrt{\frac{2}{3x+5}}}}{\sqrt[3]{14x+50}}\geq0$$ 
and we get a very big coefficients.

REPLY [2 votes]: The following proof relies on the Computer Algebra Systems (CAS).
It suffices to prove that for all $x \in [0, 1]$,
$$\sqrt[3]{\frac{1}{1+7x}}+\sqrt[3]{\frac{x}{x+7}}\geq\sqrt{\frac{1}{1+3x}}+\sqrt{\frac{x}{x+3}}.$$
We will use the following bound: for all $x\in [0, 1]$,
$$\sqrt[3]{\frac{1}{1+7x}} - \sqrt{\frac{1}{1+3x}} \ge -\frac{50x(1-x)}{659x^2+481x+60}.\tag{1}$$
The proofs are given later.
With the bound above, it suffices to prove that for all $x\in [0, 1]$,
$$\sqrt[3]{\frac{x}{x+7}} - \sqrt{\frac{x}{x+3}} \ge \frac{50x(1-x)}{659x^2+481x+60}.\tag{2}$$
If $x=0$, clearly the inequality is true. In the following, assume that $x > 0$.
With the substitution $\frac{x}{x+7} = (\frac{1}{2+v})^3$ 
(or $x = \frac{7}{v^3+6v^2+12v+7}$) for $v \ge 0$,
it suffices to prove that
\begin{align}
&\frac{60 v^6+720 v^5+3250 v^4+10047 v^3+25482 v^2+42084 v+58800}{(60 v^6+720 v^5+3600 v^4+12847 v^3+33882 v^2+50484 v+58800) (2+v)}\\
\ge\ &
\sqrt{\frac{7}{3v^3+18v^2+36v+28}}.
\end{align}
It suffices to prove that
\begin{align}
&\left(\frac{60 v^6+720 v^5+3250 v^4+10047 v^3+25482 v^2+42084 v+58800}{(60 v^6+720 v^5+3600 v^4+12847 v^3+33882 v^2+50484 v+58800) (2+v)}\right)^2\\
\ge\ &
\frac{7}{3v^3+18v^2+36v+28}
\end{align}
or
$$\frac{v^2f(v)}{(3 v^3+18 v^2+36 v+28) (2+v)^2 g(v)} \ge 0$$
where
$$g(v) = (60 v^6+720 v^5+3600 v^4+12847 v^3+33882 v^2+50484 v+58800)^2$$
and
\begin{align}
f(v) &= 10800 v^{13}+298800 v^{12}+3704400 v^{11}+28046520 v^{10}\\
&\quad +149217300 v^9+600443820 v^8+1891098627 v^7+4725531663 v^6\\
&\quad +9284097200 v^5 +13766984612 v^4+14357295264 v^3+8833795824 v^2\\
&\quad +1653512448 v+533433600.
\end{align}
We are done.
$\phantom{2}$
Proof of the bound (1): If $x = 1$, clearly the inequality in (1) is true. In the following, assume that $x < 1$.
We use the substitution $\frac{1}{1+7x} = (\frac{1+u}{2+u})^3$ (or $x = \frac{3u^2+9u+7}{7(u^3+3u^2+3u+1)}$) for $u\ge 0$.
The rest of the proof is similar to that of (2).