Question

If $x,y,z$ are three positive numbers, then the minimum value of $\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is D

$6$

Explanation for the correct option:

Finding the minimum value of $\frac{\mathbf{y}\mathbf{+}\mathbf{z}}{\mathbf{x}}\mathbf{+}\frac{\mathbf{z}\mathbf{+}\mathbf{x}}{\mathbf{y}}\mathbf{+}\frac{\mathbf{x}\mathbf{+}\mathbf{y}}{\mathbf{z}}$:

It is given that $x,y,z$ are positive numbers,

$\Rightarrow x\ge 0,y\ge 0,\mathrm{and}z\ge 0$ and therefore, $\frac{x}{y},\frac{x}{z},\frac{y}{z},\frac{y}{x},\frac{z}{y}\mathrm{and}\frac{z}{x}\ge 0$.

We know that the Arithmetic mean of positive numbers is $\ge$ their geometric mean.

Arithmetic mean $=\frac{\mathrm{Sum}\mathrm{of}n\mathrm{obsevations}}{n}$ and Geometric mean $=\sqrt[n]{\mathrm{Product}\mathrm{of}n\mathrm{observations}}$

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{x}{y}}+{\displaystyle \frac{y}{x}}}{2}\ge \sqrt{\frac{x}{y}·\frac{y}{x}} \\ \Rightarrow \frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)\ge \sqrt{1} \\ \Rightarrow \frac{x}{y}+\frac{y}{x}\ge 2...\left(1\right) \end{gathered}$$

Similarly,

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{z}{y}}+{\displaystyle \frac{y}{z}}}{2}\ge \sqrt{\frac{z}{y}·\frac{y}{z}} \\ \Rightarrow \frac{1}{2}\left(\frac{z}{y}+\frac{y}{z}\right)\ge \sqrt{1} \\ \Rightarrow \frac{z}{y}+\frac{y}{z}\ge 2...\left(2\right) \end{gathered}$$

and

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{x}{z}}+{\displaystyle \frac{z}{x}}}{2}\ge \sqrt{\frac{x}{z}·\frac{z}{x}} \\ \Rightarrow \frac{1}{2}\left(\frac{x}{z}+\frac{z}{x}\right)\ge \sqrt{1} \\ \Rightarrow \frac{x}{z}+\frac{z}{x}\ge 2...\left(3\right) \end{gathered}$$

By adding $\left(1\right),\left(2\right),\mathrm{and}\left(3\right)$

$$\begin{gathered} \frac{x}{y}+\frac{y}{x}+\frac{z}{y}+\frac{y}{z}+\frac{x}{z}+\frac{z}{x}\ge 2+2+2 \\ \Rightarrow \frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\ge 6 \\ \Rightarrow \frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\ge 6 \end{gathered}$$

Hence, the correct answer is option D.