# Compare

• Jan 6th 2011, 11:36 PM
lebanon
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a and b being two strictly positive numbers , compare (a+b)/2 and √a√b
• Jan 6th 2011, 11:59 PM
melese
Quote:

Originally Posted by lebanon
a and b being two strictly positive numbers , compare (a+b)/2 and √a√b

Consider $\displaystyle (\sqrt{a}-\sqrt{b})^2=(\sqrt{a})^2-2\sqrt{a}\sqrt{b}+(\sqrt{b})^2=a+b-2\sqrt{a}\sqrt{b}$.

Then divide by 2 to get $\displaystyle \displaystyle\frac{(\sqrt{a}-\sqrt{b})^2}{2}=\frac{a+b}{2}-\sqrt{a}\sqrt{b}$. But since $\displaystyle \displaystyle\frac{(\sqrt{a}-\sqrt{b})^2}{2}$ is nonnegative for any positive $\displaystyle a$ and $\displaystyle b$ (why?) it follows that $\displaystyle \displaystyle\frac{a+b}{2}-\sqrt{a}\sqrt{b}\geq0$.

Therefore, $\displaystyle \displaystyle\frac{a+b}{2}\geq\sqrt{a}\sqrt{b}$ for any positive numbers $\displaystyle a$ and $\displaystyle b$.
• Jan 7th 2011, 12:08 AM
TheCoffeeMachine
And it can further be proven by simple induction that:

$\displaystyle \displaystyle \left(\prod_{1\le i \le n}a_{i}\right)^{\frac{1}{n}} \le \frac{1}{n}\sum_{1\le i\le n}a_{i}$ for $\displaystyle a_{i} \in\mathbb{R}^{+}$ and $\displaystyle n\in\mathbb{N}$.