Let $a, b \in \mathbb{R}^+$. Prove that $\frac{2a^2+3b^2}{2a^3+3b^3} + \frac{2b^2+3a^2}{2b^3 + 3a^3} \le \frac{4}{a+b}$. 
Let $$a, b \in \mathbb{R}^+$$ Prove that $$\frac{2a^2+3b^2}{2a^3+3b^3} + \frac{2b^2+3a^2}{2b^3 + 3a^3} \le \frac{4}{a+b}$$

We can make the denominators common on the LHS by AM-GM but the problem is there are 2 terms and the power is 3, so I am unable to simplify it. It gives
$$5(a^2+b^2)(a+b) \le 8 \cdot 6^{3/2} \cdot (ab)^{3/2}$$
I am not able to proceed. Please help.
Thanks.
 A: Consider
$$\frac{2a^2+3b^2}{2a^3+3b^3}(a+b) = 1 + a b \frac{2a+3b}{2a^3+3b^3}$$
and from Holder, $(2a^3+3b^3)(2+3)^2 \geqslant (2a+3b)^3$, so we may write
$$LHS \times(a+b) \leqslant 2 + 25ab\left(\frac1{(2a+3b)^2} + \frac1{(2b+3a)^2}\right)$$
so it is enough to show that
$$25ab[(2b+3a)^2+(2a+3b)^2] \leqslant 2(2a+3b)^2(2b+3a)^2$$
$$\iff 59(a^2-b^2)^2 + 13(a^4+b^4-a^3b-ab^3) \geqslant 0$$
which is obvious.
A: Sometimes the most elementary and the most straightforward is the best.
It is easy to factorize $(RHS- LHS)$ as $(a-b)^2\times(something)$, isn't it?
$$\left(\frac{2}{a+b} - \frac{2a^2+3b^2}{2a^3+3b^3} \right) +\left(\frac{2}{a+b} - \frac{2b^2+3a^2}{2b^3 + 3a^3} \right) \ge 0$$
To make it more elegant:
\begin{align}
\frac{2}{a+b} - \frac{2a^2+3b^2}{2a^3+3b^3} &= \frac{2a^3+3b^3-2a^2b-3ab^2}{(a+b)(2a^3+3b^3)} \\
&=\frac{(a-b)(2a^2-3b^2)}{(a+b)(2a^3+3b^3)} \\
&=\frac{(a-b)(2a^2-2b^2)}{(a+b)(2a^3+3b^3)} - \frac{(a-b)b^2}{(a+b)(2a^3+3b^3)}\\
&\ge - \frac{(a-b)b^2}{(a+b)(2a^3+3b^3)}.
\end{align}
Similarly, 
$$\frac{2}{a+b} - \frac{2b^2+3a^2}{2b^3+3a^3} \ge - \frac{(b-a)a^2}{(a+b)(2b^3+3a^3)}.$$
Taking the sum of the two inequalities, it remains to prove
$$- \frac{(a-b)b^2}{2a^3+3b^3} - \frac{(b-a)a^2}{2b^3+3a^3} \ge 0$$
or equivalently
$$(a-b)(2a^5+3a^2b^3-2b^5-3a^3b^2) \ge 0,$$
which is true from the following factorization (again):
$$2a^5+3a^2b^3-2b^5-3a^3b^2 = 2(a^5-b^5) + 3(a^2b^3 - a^3b^2) = (a-b)(2a^4+2a^3b-a^2b^2+2ab^3+2b^4).$$
