Question

If x, y, z are three + ive real numbers, then 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

L. H. S $=(\frac{y}{z}+\frac{x}{y})+(\frac{z}{x}+\frac{x}{z})+(\frac{y}{z}+\frac{z}{y})$
Each pair is $\ge 2$ as A.M. $\ge G.M.$
$\therefore$ L.H.S $\ge 6$. Hence the minimum value is 6
Another form:
$yz(y+z-2x)+zx(z+x-2y)+xy(x+y-2z)\ge 0$
or $yz(y+z)+zx(z+x)+xy(x+y)\ge 6xyz$
or $\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}>6$ etc.