<s>[INST] Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities. 

Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that 
  \begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\end{equation}

Thanks.
 [/INST] I don't have a rearrangement inequality proof yet, but I really like the following proof I got.
First note that $a+b \ge 2 \sqrt{ab} = 2$ by AM-GM.
$a^2 + 3 = a^2 + 3ab = a(a+3b)  \ge a(2 + 2b) = 2ab(a + 1) = 2(a+1)$, and $b^2 + 3 = b^2 + 3ab = b(b+3a) \ge b(2 + 2a) = 2b(1+a)$.
Thus, we have
$$\frac{a}{a^2+3}+\frac{b}{b^2+3}  \le \frac{a}{2(a+1)} + \frac{1}{2(a+1)} = \frac{1}{2}$$
and we're through!
</s>