CLASSICAL INEQUALITY FOR AVERAGES

From the inequality  follows that , and the sign of equality takes place only when .

If  – are positive numbers, then if we divide both parts of the last inequality by , we get:

                                                                                          (4)

Using inequality (4), we can easily prove that the sum of two positive numbers is not less than 2 if their product is one.

In fact, if . The inequality  follows from inequality (4) for  and .

Let us now prove the theorem.

Theorem 1. If the product of n positive numbers is 1, then their sum is not less than n.

In other words, the equation  it follows that , where , if the numbers  are not all the same.

Proof.  Let's prove this theorem by mathematical induction.

Earlier we proved the validity of Theorem 1 for the case of two positive numbers (n=2).

Let us suppose that the theorem is true for n=k≥2, i.e. supposing that the inequality

takes place if , we prove the theorem for , i.e. we prove that

if , with .

First of all, we note that if

,

then two cases can be presented:

1. All multipliers  are equal, i.e.
2. 
2. Not all multipliers are equal.

In the first case each multiplier is equal to one and their sum is , i.e.

In the second case among the multipliers of the product  we can find the numbers that are more or less than one  (if all multipliers were  less than one then their product  would be less then one)

Let, for example, . We have:

.

Assuming , we will get

Since here is the product  of positive numbers are equal to one, then  (according to the assumption) their sum is not less than k, i.e.

But

Remember that, , we will get

Since , то  hence

So, theorem 1 is proved.

Task 1.  Prove, that  if  are  positive numbers, then

 the sign of inequality takes place in case when  .

Solution. Since

,

then the inequality follows from the theorem 1. The equal sign takes place  when

i.e. when .

Task 2. Prove the inequality

.

Solution. We have:

Since the product of the terms on the right side of the equality is equal to one, then  their sum is not less than two The equal sign takes place only when. 

Task 3. The equal sign takes place only when

Solution. Since , then

Task 4. Prove the inequality

Solution. Divide the numerator and denominator of the left side of the inequality by  

Since, , то , hence,

Definition. Number  is called geometric mean of positive numbers  and the number  is called geometric mean of these numbers.

Theorem2. The geometric mean of positive numbers is not greater than the arithmetic mean of the same numbers.

If number  are not all equal, then the geometric mean of these numbers is less than their arithmetic mean.

Proof. From equality  follows that

 Since the product of n positive numbers is 1, then (Theorem1) their sum is not less than  , i.e.

 Multiplying both parts of the last inequality by g and dividing by n, we get:

Note that equality takes place only when

, i.e. .

If the numbers  are not all equal, then

References:

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6. S. Furuichi, On refined Young inequalities and reverse inequalities, J.Math.Ineq. 5 (2011) 21–31.