For each value for $a$ and $b$, being real numbers, $a^2 + b^2 \ge ab$
Should I solve this by replacing all possible real number formulas in these to be able to prove this? As an example for odd numbers, by replacing $2x + 1$, etc...?
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2We even have $a^2+b^2\geq2ab$. That's stronger, and easier to prove. – Arthur Jul 14 '21 at 23:30
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1You have to prove this for any real numbers, so the strategy you are suggesting will not be useful as you cannot cover all types of numbers. – Anurag A Jul 14 '21 at 23:30
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1@Arthur Yes but $2ab \geq ab$ is not true in general – Stephen Donovan Jul 14 '21 at 23:30
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If it's about real numbers it doesn't make sense to speak of odd or even. But you could try to prove the stronger inequality $a^2+b^2 \ge 2 |ab|$. – dxiv Jul 14 '21 at 23:31
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1@StephenDonovan True, but the LHS is always positive, so $2ab\geq ab$ holds in any case of interest. – Arthur Jul 14 '21 at 23:31
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1@Arthur True. So we prove by cases then – Stephen Donovan Jul 14 '21 at 23:32
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Try to write it as $a^2+b^2-ab=\frac{2a^2+2b^2-2ab}{2}=\frac{a^2+b^2+(a-b)^2}{2}$. Now all three terms are $\geq 0$, so you get $a^2+b^2\geq ab$.
Anurag A
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Alternative:
Either $(a \times b)$ is negative, or it isn't.
$\underline{\text{Case 1:} ~(a \times b) ~\text{is negative}}$
Then, $a^2 + b^2 \geq 0 > (a \times b).$
$\underline{\text{Case 2:} ~(a \times b) ~\text{is not negative}}$
Then $2(a \times b) \geq (a \times b).$
Further, since $(a - b)^2 \geq 0$ you have that
$a^2 + b^2 \geq 2(a \times b) \geq (a \times b).$
user2661923
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1Stylistically ,I find it odd using $\times$ for regular multiplication, even though it had its origins there! – Alan Jul 15 '21 at 00:16
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@Alan I took literary license, thinking that $(a \times b)$ results in a more emphatic exposition than $(ab)$. – user2661923 Jul 15 '21 at 00:18
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I usually use \cdot $\cdot$ when I want to emphasize the multiplication operator – Alan Jul 15 '21 at 00:19