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It's a fact that:

$\frac{a+b}{2} \ge \sqrt{ab}$

When a and b are non negative real numbers. What's the error in the following demonstration:

$\frac{a+b}{2} \ge \sqrt{ab}$ , then

$a+b \ge 2\sqrt{ab}$ , then

$a^{2}+2ab+b^{2} \ge 4ab$ , then

$a^{2}-2ab+b^{2} \ge 0$ , then

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

discrete math professor gave that question as homework, and this is a direct translation from portuguese.

the only idea I have is to say that both $(a-b)^{2}\ge 0$ and $(b-a)²\ge 0$ are valid, but it only specifies one. Is that correct?

Hanul Jeon
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Woray
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    Welcome to MSE. Who says there’s an error? $(a-b)^2=(b-a)^2$ – J. W. Tanner Mar 18 '24 at 00:03
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    How do you "know" $(a+b)/2 > \ge \sqrt{ab}$? The usual way to prove that starts with $(a-b)^2 \ge 0$. – Ethan Bolker Mar 18 '24 at 00:07
  • Possible duplicates: https://math.stackexchange.com/q/595034/532409 https://math.stackexchange.com/q/2343337/532409 https://math.stackexchange.com/q/557653/532409 https://math.stackexchange.com/q/4614921/532409 – Quillo Mar 18 '24 at 00:18
  • Why do you think there's an error? – H. sapiens rex Mar 18 '24 at 00:27
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    There's nothing wrong with that demonstration. It just doesn't mean anything as $(a-b)^2\ge 0$ is always true =whether $\frac {a+b}2 \ge \sqrt ab$ or not. You might as well say "Fish like tacos $\implies a-b\in \mathbb R \implies (a-b)^2 \ge 0$. The thing though is EVERY step is reversible. $\frac {a+b}2\ge \sqrt{ab}\iff a+b\ge 2\sqrt{ab}\iff (a+b)^2\ge 4ab$ AND $a,b\ge 0\iff a^2+2ab +b^2 \ge 4ab;a,b\ge 0\iff a^2-2ab-b^2\ge 0;a,b\ge 0\iff (a-b)^2\ge 0;a,b\ge 0$. As $(a-b)^2\ge 0$ always and we were told $a,b\ge 0$ this means we have proven $\frac {a+b}2\ge \sqrt{ab}$. – fleablood Mar 18 '24 at 00:32
  • @fleablood. What happens if we assume the negation? $\frac{a + b}{2} < \sqrt {ab} \implies a + b < 2\sqrt {ab} \implies a^2 + b^2 + 2ab < 4ab \implies a^2 - 2ab + b^2 < 0 \implies \left(a - b \right)^2 < 0$ – Agent Smith Mar 18 '24 at 07:36
  • Then that would be a proof by contradiction because we know $(a-b)^2 \not < 0$. – fleablood Mar 18 '24 at 14:56
  • To me this issue is the word "demonstration". Was this a mistranslation of the word "proof"? A "demonstration" isn't a proof. A demonstration is just "here I showed you something". And there's nothing wrong in what is being shown. But as a proof. What's wrong is you can' prove something by assuming it is true and coming to a true conclusion. That only shows that assuming something true didn't lead to an obvious contradict. Here's a proof that it rains lemonade. Assume it rains lemonade. Then after a rain the ground would be wet. The ground is wet after a rain. So it rains lemonade. – fleablood Mar 18 '24 at 15:04
  • All that's wrong here is that "then" should read "because", i.e., the proof presents the right steps but connects them in the reverse order. – Rob Arthan Mar 18 '24 at 20:58

1 Answers1

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It's a common mistake in proofs to assume the truth of what you wish to prove. Once that has been done, you can prove $anything$.

For example, here is an absurd "proof" that $$2=3\Rightarrow 0=0\Rightarrow \text{Last statement is valid}$$

The above proof would have been correct if you begin with the last step, and work backwards.

Or, if you write at the very end that: All steps are reversible.

That last statement, of course, needs to be checked very carefully.

Starlight
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