Let . Prove that

**Solution**

The LHS is equal to which by AM – GM is less or equal to

where . Since it follows from Bernoulli inequality that .

Skip to content
# Tag: Inequality

## Nested radical inequality

## Root inequality

## Inequality with roots

## Inequality in acute triangle

## Inequality in a triangle

A site of university mathematics

Let . Prove that

**Solution**

The LHS is equal to which by AM – GM is less or equal to

where . Since it follows from Bernoulli inequality that .

Let be positive real numbers such that . Prove that

**Solution**

Well if we apply AM-GM to we obtain

(1)

and similarly if we apply AM – GM to we obtain

(2)

We have successively,

Let be positive real numbers. Prove that

**Solution**

We apply the AM – GM inequality, thus:

Hence it suffices to prove that which holds because it is equivalent to .

Let be an acute triangle. Prove the following inequality

**Solution**

The solution can be found at cut -the – knot.

Given a triangle let denote the median points of the sides respectively. Prove that

where denotes the the circumradius and the inradius respectively.

*(Adil Abdullayev / RMM)*

**Solution **[Soumava Chakraborty]

We have successively

In all the above denotes the semiperimeter of the triangle. More on Gerretsen’s inequality can be found at this link .