If $a$ and $b$ are positive real numbers, then $a+ b \ge 2 \sqrt{ab}$

Question

If $a$ and $b$ are positive real numbers, then $a+ b \ge 2 \sqrt{ab}$.

Can you guys help me with this problem? I'm still trying to learn how to do proving and I don't really know how to start or complete this problem.

Answer 1

\begin{align} 0 & \le (a-b)^2\\ &=(a+b)^2 -4ab\\ 4ab &\le (a+b)^2 \end{align} Taking square root preserves inequality since both $a+b$ and $2\sqrt{ab}$ are positive

Answer 2

$$(a+b-2\sqrt {ab})=(\sqrt a-\sqrt b)^2 \ge 0$$

Thus $$ a+b \ge 2\sqrt {ab}$$

Answer 3

$$a+b <= 2 \sqrt (ab) $$ is incorrect for positive a,b.

The equation is valid for negative a,b

Which, ofcourse is clearly visible 'a negative number is less than a positive number on number line.'

But for positive a,b we have the inequality $$a+b>=2\sqrt{ab} $$

Hint: let $$\sqrt{a}= x $$ and $$\sqrt{b}=y $$, it is valid since a,b are positive.

Now you must be able figure out a whole term which is necessarily greater than equal to 0.