Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ [duplicate]

Solution 1:

Depends on the machinery you are allowed. It is immediate from Cauchy-Schwarz.

Solution 2:

There is a basic inequality you should know:

$$(a_{1}+a_{2}+\cdots+a_{n}) \cdot \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots +\frac{1}{a_{n}}\right) \ge n^2 $$ that is elementarily solved by Cauchy-Schwarz in a single row proof $$\left(\sqrt{a_{1}}\frac{1}{\sqrt{a_{1}}}+\cdots +\sqrt{a_{n}}\frac{1}{\sqrt{a_{n}}}\right)^2=n^2\le (a_{1}+\cdots+a_{n}) \cdot \left(\frac{1}{a_{1}}+\cdots +\frac{1}{a_{n}}\right) $$

Q.E.D.

Solution 3:

It's a convexity inequality: the function $f(x) = \frac{1}{x}$ is convex on $\mathbb{R}^{+*}$ (its second derivative is positive), so for any $a,b,c\in\mathbb{R}^{+*}, \frac{1}{3}(f(a)+f(b)+f(c))\geq f(\frac{a+b+c}{3})$, which is $\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq\frac{3}{a+b+c}$, or finally $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{9}{a+b+c}$