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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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