Proving inequality involving radical I have tried many examples for the following inequality in Mathematica. It is likely true. I need some help proving it. 
For $x_1, x_2, y_1, y_2 \geq  0$, 
\begin{align*}
6(x_1^5y_1 + x_1y_1^5) + 4(x_2^3y_2 + x_2y_2^3) \leq 6(x_1^6 + x_2^4)^{\frac{5}{6}}(y_1^6 + y_2^4)^{\frac{1}{6}} + 6(x_1^6 + x_2^4)^{\frac{1}{6}}(y_1^6 + y_2^4)^{\frac{5}{6}}.
\end{align*}
I guess the difficulty of this is from $x_2$ and $y_2$ parts because the highest and lowest degrees of them from LHS and RHS are not the same. 

Note by Hölder's inequality
$$(x_1^6+x_2^4)^{5/6} (y_1^6+y_2^4)^{1/6} \geqslant x_1^5y_1+x_2^{10/3}y_2^{2/3}$$
and
$$(x_1^6+x_2^4)^{1/6} (y_1^6+y_2^4)^{5/6} \geqslant x_1y_1^5+x_2^{2/3}y_2^{10/3}$$
Using this, it is left to show
$$3(x_2^{10/3}y_2^{2/3}+x_2^{2/3}y_2^{10/3}) \geqslant 2(x_2^3y_2 + x_2y_2^3)$$
In fact as $(\frac{10}3, \frac23) \succ (3, 1)$, by Muirhead (or you can use AM-GM) the stronger inequality holds:
$$x_2^{10/3}y_2^{2/3}+x_2^{2/3}y_2^{10/3} \geqslant x_2^3y_2 + x_2y_2^3$$
Hence:
$$6(x_1^5y_1 + x_1y_1^5) + \color{red}{6}(x_2^3y_2 + x_2y_2^3) \leq 6(x_1^6 + x_2^4)^{\frac{5}{6}}(y_1^6 + y_2^4)^{\frac{1}{6}} + 6(x_1^6 + x_2^4)^{\frac{1}{6}}(y_1^6 + y_2^4)^{\frac{5}{6}}$$
