Question

If $x+y+z=1$, show that the least value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $9$; and that $(1-x)(1-y)(1-z)>8xyz$.

Answer 1

According to power mean inequality ,

$QM\ge AM\ge GM\ge HM$

Therfore we can say that arithematic mean is greater than harmonic mean .

$\frac{x+y+z}{3}\ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

As , $x+y+z=1$

$\frac{1}{3}\ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

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

$9\le \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

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

Thus least value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $9$

When $AM\ge GM$

$\frac{x+y}{2}\ge \sqrt{xy}$

$\frac{x+z}{2}\ge \sqrt{xz}$

$\frac{y+z}{2}\ge \sqrt{yz}$

As $x+y+z=1$

$x+y=1-z$

$x+z=1-y$

$y+z=1-x$

$\frac{x+y}{2}=\frac{1-z}{2}\ge \sqrt{xy}$

$\frac{x+z}{2}=\frac{1-y}{2}\ge \sqrt{xz}$

$\frac{y+z}{2}=\frac{1-x}{2}\ge \sqrt{yz}$

Multiply the above 3 equation we get ,$(1-x)(1-y)(1-z)\ge 8\sqrt{x^2{y}^2{z}^2}$

$(1-x)(1-y)(1-z)\ge 8\sqrt{xyz}$