Question

If $x,y\mathrm{and}z$ are three positive real numbers, then minimum values 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.

We know that for all positive real numbers, $A.M\ge G.M$.

So, for $\frac{y}{x},\frac{z}{x},\frac{z}{y},\frac{x}{y},\frac{x}{z},\frac{y}{z}$

$\begin{array}{rcl}& \Rightarrow & \frac{\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}}{6}\ge {\left(\frac{y}{x}\times \frac{z}{x}\times \frac{z}{y}\times \frac{x}{y}\times \frac{x}{z}\times \frac{y}{z}\right)}^{1/6}\\ \Rightarrow \frac{\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}}{6}& \ge & 1\\ & \Rightarrow & \frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\ge 6\\ \Rightarrow \frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}& \ge & 6\end{array}$

So, the minimum value of $\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}$ is $6$.

Hence, option D is correct.