Question

Given that $P\ge 1$. What is the minimum value of $P^2+\frac{625}{P^2}$ ?

• A. 10
• B. 25
• C. 50
• D. 75

Answer 1

The correct option is C 50

As we know that A.M $\ge$ G.M.

Let two numbers are $P^2$ and $\frac{625}{P^2}$.

A.M of these two numbers $=\frac{P^2+\frac{625}{P^2}}{2}.$

G.M of two given two numbers $={(P^2\times \frac{625}{P^2})}^{\frac{1}{2}}$

$\frac{P^2+\frac{625}{P^2}}{2}\ge {(P^2\times \frac{625}{P^2})}^{\frac{1}{2}}\Rightarrow P^2+\frac{625}{P^2}\ge 50.$

So, the minimum value of given expression is 50.