<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] We can assume $a \leq 1 \leq b$. Applying rearrangement inequalities to
$$
\begin{align}
a &\leq 1 \\
1 &\leq b
\end{align}
$$
we get 
$$
a + b \geq 1 + ab = 2
$$
and
$$
b + 3a \geq 2a + 2 \\
a + 3b \geq 2b + 2
$$
Therefore
$$
\begin{align}
\frac{a}{a^2+3}+\frac{b}{b^2+3} &= \frac{1}{a + 3b} + \frac{1}{b + 3a} \leq\\ &\frac{1}{2b + 2} + \frac{1}{2a + 2} = \frac{a}{2a + 2} + \frac{1}{2a + 2} = \frac 1 2
\end{align}
$$
</s>