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.
 A: 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
$$
\begin{align}
\frac{a}{a^2+3}+\frac{b}{b^2+3} &= \frac{1}{a + 3b} + \frac{1}{b + 3a} \leq\\ &\frac{1}{2b + 2} + \frac{1}{2a + 2} = \frac{a}{2a + 2} + \frac{1}{2a + 2} = \frac 1 2
\end{align}
$$
A: 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!
A: Since $ab=1$, we have $b = \frac{1}{a}$, and thus your inequality is equivalent to $\displaystyle\frac{a}{a^2+3}+\frac{a}{1+3a^2}\leq\frac{1}{2}$.
You can simply define $f(a) = \displaystyle\frac{a}{a^2+3}+\frac{a}{1+3a^2}$ and maximize $f$. 
A: Homogenize the given problem into,
$$\frac{\sqrt{ab}}{a+3b}+\frac{\sqrt{ab}}{b+3a}\leq \frac 12.$$
Now note that, using the AM-GM inequality we have $a+3b\geq 2\sqrt{2b(a+b)},$ so that $$\dfrac{\sqrt{ab}}{a+3b}\leq \frac{\sqrt{a}}{2\sqrt{2(a+b)}}.$$
Hence it suffices to check that $\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{2(a+b)}}\leq 1,$ which is perfectly true from the Cauchy-Schwarz inequality.
$\Box$
A: An opportunity to show the defects of AM-GM, you get a weaker inequailty.
$\dfrac{a^2+3}{a}=a+\dfrac{3}{a} \ge 2 \sqrt{3} \implies (\dfrac{a^2+3}{a})^{-1} \le \dfrac{1}{2 \sqrt{3}}$
$\dfrac{b^2+3}{b}=b+\dfrac{3}{b} \ge 2 \sqrt{3} \implies (\dfrac{b^2+3}{b})^{-1} \le \dfrac{1}{2 \sqrt{3}}$
When you add them, you have weaker inequality. Therefore, re-arrangement inequality is an appropriate one!
