If $a,b,c$ are positive, then $(a+b+c)(1/a+1/b+1/c)\ge 9$

Question

The question asks to prove that if "$x_1,x_2,x_3$ are positive numbers show that: $$(x_1+x_2+x_3) \left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3} \right)\ge 9$$

I've tried to use the fact that the arithmetic mean is greater than the harmonic mean since it looks like it's kind of in that form but it doesn't seem to work. I tried expanding it too and simplifying a little to get: $$ x_1^2(x_2+x_3)+x_2^2(x_1+x_3)+x_3^2(x_1+x_2)\ge 6 x_1x_2x_3$$

but I can't seem to see anyway to further get it into a form that easily proves the required inequality. Any hints as to a way to do this?

Answer 1

The Arithmetic-harmonic mean inequality is actually the usual way of doing this.

We have $ \frac{x_1+\cdots+x_n}{n}\geq \frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}} $ in general. Write it out for $n=3$, and stare at it for a bit - it's a very simple manipulation.

Answer 2

As $x\to 1/x$ is convex on $(0,\infty)$:

$$ 1/\left[ \frac 13 \left( x+y+z\right)\right] \le \frac 13 \left( 1/x+1/y+1/z\right) \\ 1\le \frac 13 \left( 1/x+1/y+1/z\right)\frac 13 \left( x+y+z\right) \\ 9\le \left( 1/x+1/y+1/z\right)\left( x+y+z\right) $$

Answer 3

You use the Cauchy inequality, $x+y+z\ge 3\sqrt[3]{xyz}$ for all $x,y,z\ge 0$, to prove the statement. Solve: We have $$a+b+c\ge 3\sqrt[3]{abc},$$ $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3\sqrt[3]{{\frac{1}{a}.\frac{1}{b}.\frac{1}{c}}} = \frac{3}{{\sqrt[3]{{abc}}}}.$$ So, we have $$\left( {a + b + c} \right)\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \geqslant 3\sqrt[3]{{abc}}.\frac{3}{{\sqrt[3]{{abc}}}} = 9.$$ The inequality is proved.

Answer 4

Let $u=\langle\sqrt{a},\sqrt{b},\sqrt{c}\rangle$ and $v=\langle\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}}\rangle$.

By the Cauchy-Schwarz inequality,

$(u\cdot v)^2\le|u|^2|v|^2$ so $\displaystyle9\le(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.