*Problem:*

Let be positive real numbers such that . Prove that .

*Problem:*

Let be positive real numbers such that . Prove that .

*Problem:*

Let be positive real numbers. Prove that

.

*Solution (An idea by Silouanos Brazitikos):*

From the above inequality is it enough to show that

.

But from Schur’s Inequality we know that

.

So it is enough to check that

.

From Cauchy-Schwarz Inequality we know that

.

Therefore we only need to prove that

,

which is obviously true from AM-GM Inequality, *Q.E.D.*

*Problem:*

If are positive real numbers satisfying the equality , then prove that

.

*Solution:*

Without loss of generality, assume that .

Since

and

,

using Chebyshev’s Inequality we get:

.

Moreover, From the AM-GM Inequality we have that

.

Therefore, it suffices to prove that

.

Set .

The above inequality can be written now as

, or .

From the AM-GM Inequality once again, we acquire that

.

And thus, it is enough to check that

,

which is equivalent to the obvious inequality

, *Q.E.D.*

*Problem:*

If are positive real numbers such that , then prove that

.

*Solution:*

From the Cauchy-Schwarz inequality we deduce that

.

Removing the square we get that

. Let us now divide by .

Then we have .

Moreover, multiply by . We, thus, acquire

.

So,we have proved that

.

We will now apply Holder’s Inequality, that is

, or .

Rewrite the sum as .

Then we get that:

.

And finally, from the AM-GM inequality we have

So, we have proved that

,* Q.E.D.
*

*Problem:*

Let be positive real numbers such that . Prove that

.

*Solution:*

From the hypothesis, the inequality is of the form . If we expand the nominators and the denominators, then we get that

.

Now let us use once again the Cauchy-Reverse technique.

.

Thus, we have

.

Moreover,

.

So,

.

Multiplying by we acquire

.

Now, if we sum up the inequalities it remains to prove that

,

which reduces to the obvious inequality , *Q.E.D.*

*Problem:*

If prove that

.

*Solution:*

From the well-known lemma we deduce that:

.

Doing that cyclic for and adding up the relations we get that

.

So, it is enough to prove that

.

Squaring both sides we come to the conclusion

, *Q.E.D.*