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

Please help me for prove this inequality: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$

## marked as duplicate by Watson, Willie Wong, Daniel W. Farlow, Alexis Olson, астон вілла олоф мэллбэргNov 11 '16 at 0:01

This inequality can also be rewritten as

$$\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \,,$$

which is just the AM-HM inequality.

A more direct proof would be to simply multiply:

$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3+ (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{c}{b}+\frac{b}{c}) \geq 3+2+2+2 =9$$

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

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.

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 $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{9}{a+b+c}$

Inequality can be written as:

$$\left(a+b+c\right) \cdot \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9 .$$

And now we apply the $AM-GM$ inequality for both of parenthesis. So:

$\displaystyle \frac{a+b+c}{3} \geq \sqrt[3]{abc} \tag{1}$ and $\displaystyle \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \geq \frac{1}{\sqrt[3]{abc}} \tag{2}.$ Now multiplying relation $(1)$ with relation $(2)$ we obtained that : $$\left(\frac{a+b+c}{3}\right) \cdot \left(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\right) \geq \frac{\sqrt[3]{abc}}{\sqrt[3]{abc}}=1.$$ So, we obtained our inequality.

Beferore we prove this inequality first prove that the following inequality:

For $a>0$, $b>0$ is true this inequality: $\frac{a}{b}+\frac{b}{a}\geq 2$

$\frac{a}{b}+\frac{b}{a}-2$=$\frac{a^2+b^2-2ab}{ab}$=$\frac{(a-b)^2}{ab}$.

Since $a>0$, $b>0$ $\Rightarrow$ $a-b>0$ $\Rightarrow$ $(a-b)^2>0$, $ab>0$.

Means, $(a-b)^2>0$, $ab>0$ $\Rightarrow$ $\frac{(a-b)^2}{ab}>0$.

Since $a=b$ $\Rightarrow$ $\frac{(a-b)^2}{ab}=0$.

Means, $\frac{(a-b)^2}{ab}\geq 0$ $\Rightarrow$ $\frac{a}{b}+\frac{b}{a}-2$=$\frac{(a-b)^2}{ab}\geq 0$.

$\frac{a}{b}+\frac{b}{a}-2\geq 0$ $\Rightarrow$ $\frac{a}{b}+\frac{b}{a}\geq 2$.

Now turn to prove the first inequality.

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}$

Since $a>0$, $b>0$, $c>0$ $\Rightarrow$ $a+b+c\geq 0$, so

$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c)$=$(\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3\geq 2+2+2+3=9$