Prove that the sum of two positive real numbers is equal or greater than the square root of their product.

Question

Trying to prove this:

$A$ and $B$ are positive real numbers.

$A + B \geq \sqrt{AB}$

This is what I wrote:

Proof by Contradiction

$A + B < \sqrt{AB}$

$(A + B)^2 < AB$

$A^2 + AB + AB + B^2 < AB$

$A^2 + AB + B^2 < 0$

Inconsistent with $A>0$ and $B>0$.

Did I do this correctly?

Try not to post questions that can be answered with yes/no. Do you have doubts about a specific part? — Paramar, Apr 27 '15 at 00:09
Your proof looks good. This is kind of a special case of the AM/GM inequality. $\frac{a_1+a_2+...+a_{n}}{n}\geq \sqrt[n]{a_1\cdot a_2\cdot ...\cdot a_{n}}$ with $n=2$. — TravisJ, Apr 27 '15 at 00:13

score 1 · Accepted Answer · answered Apr 27 '15 at 00:14

1

Yes, it's correct.

You could have proceeded directly, obtaining $$ A^2+AB+B^2\geq0 $$ which is true when $A,B>0$.

In fact one even has $$ \frac{A+B}{2}\geq\sqrt{AB} $$ and you might want to take a look at the AM-GM inequality.

answered Apr 27 '15 at 00:14

Guest

score 1 · Answer 2 · answered Apr 27 '15 at 00:15

1

It looks good, but contradiction isn't needed:

$AB\leq A^2+2AB+B^2 \implies \sqrt{AB} \leq A+B$.

answered Apr 27 '15 at 00:15

TorsionSquid

2 Answers2