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

(d) 6

x > 0, y > 0, z > 0

⇒ \(\frac{x}{y},\frac{y}{x},\frac{y}z,\frac{z}y,\frac{x}z,\frac{z}x\) are all positive numbers. 

∴ Applying AM – GM inequality, we have

⇒ \(\frac12\big(\frac{x}{y}+\frac{y}{x}\big)≥\big(\frac{x}{y}.\frac{y}{x}\big)^{\frac12}\)  ⇒ \(\frac{x}{y}+\frac{y}{x}≥2\)

Similarly. \(\frac{x}{y}+\frac{y}{x}≥2\), \(\frac{x}{z}+\frac{z}{x}≥2\)

∴ The minimum value of \(\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z} \) is 6.