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.
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.
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} $
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!
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 $f$.
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. $\Box$
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!