# Proving that $\sqrt{ab} \leq \dfrac{a + b}{2}$ via contradiction

Today I was trying to prove via contradiction the following:

$$\sqrt{ab} \leq \dfrac{a + b}{2}, a,b \geq 0$$

So what I did was the following:

Let's assume that $$\sqrt{ab} > \frac{a + b}{2}$$, this means: $$2\sqrt{ab} > a + b \Leftrightarrow 0 >a - 2\sqrt{ab} + b$$

We have that $$a - 2\sqrt{ab} + b = (\sqrt a - \sqrt b)^2$$ or $$a - 2\sqrt{ab} + b = (\sqrt b -\sqrt a)^2$$

So if we sub this in the expression above we end up with:

$$0 > (\sqrt a - \sqrt b)^2 \vee 0 > (\sqrt b -\sqrt a)^2$$

This is clearly a contradiction because the square of a real number is allays positive. This means that:

$$\sqrt{ab} \leq \dfrac{a + b}{2}$$

My question is: Is this correct? Because we can only apply proofs by contradictions when there is only two different answers, so if it's not one, it has to be the other. But in this case, although is seems to exists only two alternatives: $$>$$ or $$\leq$$, I think That there are 3 of them: $$>, =$$ and $$<$$ so I'm not sure whether this proof is valid or not.

• This is correct, because $\leq$ contains the two cases $=$ and $<$. But I would say that proving this inequality by contradiction is not the most natural way ! Sep 17 '20 at 20:42
• @Eduardo A minor point is that since $(\sqrt{a} - \sqrt{b})^2 = (\sqrt{b} - \sqrt{a})^2$ for all real, non-negative $a$ and $b$, you don't need to specify & handle those $2$ cases. Sep 17 '20 at 20:45
• Let assume $a=b=1 \implies 1>1$
– user
Sep 17 '20 at 20:48
• You can rewrite this as a direct proof without contradiction. It would start by assuming that $a,b \ge 0$, and then continuing by proving a sequence of equivalences: "$\sqrt{ab} < \frac{a+b}{2}$ is equivalent to $2 \sqrt{ab} < a+b$ which is equivalent to ..." Sep 17 '20 at 20:56

Let's try doing it a different way. Notice that for all real numbers $$z$$, we have $$z^2 \geq 0$$. Let's take $$a, b \geq 0$$. We have $$(\sqrt{a}-\sqrt{b})^2 \geq 0,$$ or $$a - 2\sqrt{ab} + b \geq 0$$. Now rearrange things to get $$\frac{a+b}{2} \geq \sqrt{ab}.$$