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.

  • 4
    $\begingroup$ 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 ! $\endgroup$ Sep 17 '20 at 20:42
  • $\begingroup$ @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. $\endgroup$ Sep 17 '20 at 20:45
  • $\begingroup$ Let assume $a=b=1 \implies 1>1$ $\endgroup$
    – user
    Sep 17 '20 at 20:48
  • $\begingroup$ 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 ..." $\endgroup$
    – Lee Mosher
    Sep 17 '20 at 20:56

The Silver Doe is correct, this seems fine to me.

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 $$\frac{a+b}{2} \geq \sqrt{ab}.$$

  • $\begingroup$ Notice it's the same kind of argument you do, but I didn't have to do a contradiction at any point. $\endgroup$
    – User203940
    Sep 17 '20 at 20:47
  • $\begingroup$ But what's wrong with the contradiction? Is it not elegant to use it in a proof like this? $\endgroup$ Sep 17 '20 at 20:50
  • $\begingroup$ Contradiction always works. However using contradiction sometimes leads to messy arguments. There are some people who are adamantly against it's use unless otherwise necessary. I personally don't care. $\endgroup$
    – User203940
    Sep 17 '20 at 20:51
  • $\begingroup$ @EduardoMagalhães I belong to the people who care, for several reasons :) First, I think it is an excellent training, when you make a contradiction proof, to think about if the contradiction is necessary, and if you can make about without contradiction. Secondly, the fact that contradiction proofs work is really a strong thing logically speaking, and should not be used if not necessary (the same as using a very strong theorem to prove something elementary !). Finally, contradiction proofs are often less clear, and can lead to mistakes easily : one should always consider a direct approach. $\endgroup$ Sep 17 '20 at 20:58
  • 2
    $\begingroup$ But, it is just a "stylistic" point of view. Your proof is correct, and it is the only thing that really matters ;) $\endgroup$ Sep 17 '20 at 20:59

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