$\frac{1}{a_3+a_4}+\frac{1}{a_1+a_2}\ge \frac{4}{a_1+a_2+a_3+a_4} $

Question

Prove that $$\frac{1}{a_3+a_4}+\frac{1}{a_1+a_2}\ge \frac{4}{a_1+a_2+a_3+a_4} $$

I am supposed to use AM-HM but I don't how I should. By AM-HM

We have $$\frac{1}{a_3+a_4}+\frac{1}{a_1+a_2}\ge \frac{2}{a_1+a_2+a_3+a_4} $$

Did I do something wrong?

Answer 1

Hint : Let $a_1+a_2=x$ and $a_3+a_4=y$. It remains to prove, $$\frac{1}{x}+\frac{1}{y}\ge \frac{4}{x+y}\;\; \Leftrightarrow\;\; (x+y)\left(\frac{1}{x}+\frac{1}{y}\right)\ge 4$$ The A.M-H.M inequality for two terms, $$\frac{x+y}{2}\ge \dfrac{1}{\dfrac{\frac{1}{x}+\frac{1}{y}}{2}}$$

Answer 2

First thing, you need some constraints on $a_1,a_2,a_3,a_4$. It looks like you probably intended for $a_i\in\mathbb{R}^+$ (a counter example for $a_i\in\mathbb{R}$ could be $(1,1,1,-2)$)

Since $a_i\in\mathbb{R}^+$, we have that $a_1+a_2$ and $a_3+a_4$ are both positive reals, which implies that $\frac{1}{a_1+a_2}$ and $\frac{1}{a_3+a_4}$ are both positive reals

Applying AM-HM on $\frac{1}{a_1+a_2}$ and $\frac{1}{a_3+a_4}$, we get that $$\frac{\frac{1}{a_1+a_2}+\frac{1}{a_3+a_4}}{2}\geq \frac{2}{a_1+a_2+a_3+a_4}$$ $$\frac{1}{a_1+a_2}+\frac{1}{a_3+a_4}\geq \frac{4}{a_1+a_2+a_3+a_4}$$ as required.