If $x,y$ are positive, then $\frac1x+\frac1y\ge \frac4{x+y}$

Question

For $x$, $y$ $\in R^+$, prove that $$\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}$$

Could someone please help me with this inequality problem? I have tried to use the AM-GM inequality but I must be doing something wrong. I think it can be solved with the AM-GM but I can’t solve it. Thanks in advance for your help.

Answer 1

Here is a solution with AM-GM:

$$\frac{1}{x}+\frac{1}{y} \geq \frac{2}{\sqrt{xy}}$$ $$x+y \geq 2 \sqrt{xy} \Rightarrow \frac{1}{\sqrt{xy}} \geq \frac{2}{x+y}\Rightarrow \frac{2}{\sqrt{xy}} \geq \frac{4}{x+y}$$

Also you can note that

$$(x+y)(\frac{1}{x}+\frac{1}{y}) \geq 4$$ is just Cauchy-Schwarz.

Answer 2

Hint: Your inequality is equivalent to $(x-y)^2\ge 0$, which is obviously true.

Answer 3

Without words:

$$\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\iff \frac{x+y}{xy}\ge \frac{4}{x+y}\iff (x+y)^2\ge4xy\iff (x-y)^2\ge 0$$

Answer 4

You can also use Chebyshev's inequality because if $x\geq y$ then $\frac{1}{x} \leq \frac{1}{y}$. So, we have: $$(x+y)(\frac{1}{x}+\frac{1}{y})\geq2(x\cdot\frac{1}{x}+y\cdot\frac{1}{y})=4$$