Inequality Problem $\frac 1x + \frac 1y + \frac1z \ge \frac{9}{x+y+z}$

Question

Could I please have help solving this inequality. I have tried but haven't made meaningful progress.

If x,y and z are real numbers prove that: $$\frac 1x + \frac 1y + \frac1z \ge \frac{9}{x+y+z}$$

Answer 1

Multiply your inequality by $xyz(x+y+z)$, you get : $$y^2z+yz^2+x^2z+xz^2+x^2y+xy^2\ge 6xyz$$ Now redivide by $xyz$ (maybe not a good idea to multiply initially :-) : $$\frac yx+\frac zx+\frac xy+\frac zy+\frac xz+\frac yz \ge 6$$ As it is well know that for all $u>0$, $u+\frac1u\ge2$, you get your inequality for $x,y,z>0$.

Now for the case $x,y,z$ non null reals, I don't know...

Answer 2

it must be $$x,y,z>0$$ and this inequality is equivalent to $$\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\geq 6$$ which is true. or we use $AM-GM$ twice: $$x+y+z\ge 3\sqrt[3]{xyz}$$ $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 3\sqrt[3]{\frac{1}{xyz}}$$ and multiplying both we get the statement

Answer 3

Obviously we must assume that the numbers are positive, as otherwise the inequality doesn't always hold. Nevertheless you can prove this inequality by using Cauchy-Schwartz Inequality. We have:

$$\left( \frac 1x + \frac 1y + \frac 1z \right)(x+y+z) \ge \left( \sqrt{\frac xx} + \sqrt{\frac yy} + \sqrt{\frac zz}\right)^2 = 9$$

Now dividing by $(x+y+z)$ we get the wanted inequality.

Answer 4

The claim is a striaghtforward consequence of Titu's lemma:

$$ \frac{1^2}{x}+\frac{1^2}{y}+\frac{1^2}{z} \geq \frac{(1+1+1)^2}{x+y+z}.$$