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:

\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 \\

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?


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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.

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
$$\Rightarrow a+b=\sqrt {(a-b)^2+4p}$$
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

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.