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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