# Show that $\frac{a+b}{2} \ge \sqrt{ab}$ for $0 \lt a \le b$

I have to prove that

$$\frac{a+b}{2} \ge \sqrt{ab} \quad \text{for} \quad 0 \lt a \le b$$

The main issue I am having is determining when the proof is complete (mind you, this is my first time). So I did the following steps:

\begin{align} \frac{a+b}{2} &\ge \sqrt{ab} \\ \left(\frac{a+b}{2}\right)^2 &\ge \left(\sqrt{ab}\right)^2 \\ \frac{a^2+2ab+b^2}{4} &\ge ab \\ a^2+2ab+b^2 &\ge 4ab \\ a^2-2ab+b^2 &\ge 0\\ (a-b)^2 &\ge 0 \\ \end{align}

Now this is where I stopped because if I square root each side, I will be left with $a-b \ge 0$ or in other words, $a \ge b$ which doesn't make a whole lot of sense to me. So ultimately the question is: how do I know when I'm done? and is what I did above correct?

Thanks!

## 4 Answers

You're working backward, but all of your steps are actually equivalent, so that's okay. When you take the square root, though, you get $|a-b|\ge 0.$ This is true, so you're fine.

I would start with a true statement like $|a-b|\ge 0,$ and proceed through these steps in reverse, to prove the desired inequality. Or, more simply, note that $(a-b)^2\ge 0$ for all real $a,b$, so we can get there even more quickly starting from that point.

• This might a dumb question but how would you know that you have to start of with $|a-b| \ge 0$? – Jeel Shah Nov 9 '13 at 5:25
• @gekkostate By doing the work that you have done above! You can think of what you have done as scratch work, and then write your final solution starting with $(a-b)^2 \ge 0$. It may seem strange because you are starting with something out of the blue, but that's ok. – angryavian Nov 9 '13 at 5:27
• Ohh okay. So once I have done that, in my final copy, I would write something like "It must be shown that $(a-b)^2 \ge 0$" or something like that and then I show my proof right? – Jeel Shah Nov 9 '13 at 5:29
• @gekkostate: Not quite. Instead, say something like "Since $(a-b)^2\ge0$ for all real $a,b,$ then $a^2-2ab+b^2\ge0,$ so $a^2+2ab+b^2\ge4ab,$ so...." You need to remember what it is that you're trying to prove. Also, don't forget to use the fact that $a,b$ are non-negative. That will be important in one of your steps. (Do you see where?) – Cameron Buie Nov 9 '13 at 12:55
• Ahh, Okay. I thought I had to say it at the start but I get it now and Is see where the non negative part comes in handy. Thanks! – Jeel Shah Nov 9 '13 at 14:32

The AM-GN inequality is one of the simplest and still most frequently used inequalities in elementary mathematics. Formally, it asserts the superiority of arithmetic mean over geometric mean. In symbols, the inequality says that for all real numbers $a1,a2,...,a_n$, $$\frac {a_1+a_2+...+a_n}{n} \ge(a_1a_2...a_n)^\frac{1}{n}$$ Let us claim that among all pairs of positive numbers a and b whose product is a constant,say p,the sum is minimum when the two numbers are equal,i.e,a=b=$\sqrt p$. To see this,mereley write $$a+b=\sqrt{a^2+b^2+2ab}$$ $$\Rightarrow a+b=\sqrt {(a-b)^2+4p}$$ Clearly a+b is minimum when $a-b=0$ or $a=b$ and the minimum value is $2\sqrt p$. It follows that whenever ab=p,we must have a+b$\ge$2$\sqrt p$. But this is the $A.M-G.M$ inequality for 2 real numbers.

Visualise $(\sqrt{a}-\sqrt{b})^2\ge 0$

Expand and rearrange and you will be left with your problem

• It must be $(\sqrt a-\sqrt b)^2\ge0.$ – Harsh Kumar Jul 1 '17 at 4:00
• Yes @harsh edited – Atul Mishra Jul 1 '17 at 9:19

Actually, you don't need to square both sides, since $\frac{a+b}{2} \ge \sqrt{ab} \Leftrightarrow \frac{a+b-2\sqrt{ab}}{2} \ge 0 \Leftrightarrow \frac{(\sqrt a - \sqrt b)^2}{2} \ge 0$. And for any real number x, $x^2 \ge 0$ is always true, so $\frac{(\sqrt a - \sqrt b)^2}{2} \ge 0$ is true.

As for your last equation, $(a-b)^2 \ge 0$, for any real number $a, b$, it is always a truth.