How-may-I-prove-the-following-theorem-a-b-2-ab-Thank-you-

Question Number 47476 by hassentimol last updated on 10/Nov/18

How may I prove the following theorem ?

\frac{\boldsymbol a + \boldsymbol b}{2} ⩾ \sqrt{\boldsymbol a b}

Thank you

Commented by prakash jain last updated on 11/Nov/18

This is true only for a, b ⩾ 0

Answered by Joel578 last updated on 11/Nov/18

Assume a, b \in R, then \sqrt{a}, \sqrt{b} \in R

Observe that for all \sqrt{a}, \sqrt{b} \in R

{(\sqrt{a} - \sqrt{b})}^{2} ⩾ 0

\Leftrightarrow a - 2 \sqrt{a b} + b ⩾ 0

\Leftrightarrow a + b ⩾ 2 \sqrt{a b}

\Leftrightarrow \frac{a + b}{2} ⩾ \sqrt{a b}

Hence, proved

Commented by hassentimol last updated on 11/Nov/18

Thank you sir!

It is also very helpful!

Answered by …. last updated on 10/Nov/18

since {(a - b)}^{2} ⩾ 0

\Rightarrow a^{2} + b^{2} - 2 ab ⩾ 0

\Rightarrow a^{2} + b^{2} - 2 ab + 4 ab ⩾ 0 + 4 ab

\Rightarrow a^{2} + b^{2} + 2 ab ⩾ 4 ab

\Rightarrow {(a + b)}^{2} ⩾ 4 ab

\Rightarrow (a + b) ⩾ 2 \sqrt{ab}

\Rightarrow \frac{a + b}{2} ⩾ \sqrt{ab}

\frac{\frac{}{}}{}

Commented by hassentimol last updated on 11/Nov/18

Thank you sir .

It is very helpful!