# Inequalities – Proving inequality of means for n=2 – Exercise 1904

Exercise

Prove the inequality:

$$\sqrt{ab}\leq \frac{a+b}{2}$$

Note: this inequality is the n=2 case in Inequality of arithmetic and geometric means.

$${(a_1\cdot a_2 \cdot ...\cdot a_n)}^{\frac{1}{n}}\leq\frac{a_1+a_2+...+a_n}{n}$$

Proof

We use short multiplication furmula:

$${(a-b)}^2=a^2-2ab+b^2$$

The following sentence is true for all a and b:

$$0\leq {(a-b)}^2$$

Since every squared expression is positive.

Open the brackets:

$$0\leq a^2-2ab+b^2$$

We express the middle expression as a sum and we get:

$$0\leq a^2-4ab+2ab+b^2$$

Rearrange:

$$0\leq a^2+2ab+b^2-4ab$$

We use short multiplication furmula again:

$$0\leq {(a+b)}^2-4ab$$

$$4ab\leq {(a+b)}^2$$

$$ab\leq \frac{{(a+b)}^2}{4}$$

$$\sqrt{ab} \leq \frac{a+b}{2}$$

And we got the desired result.

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