<s>[INST] For each value for $a$ and $b$, being real numbers, $a^2 + b^2 \ge ab$ 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...?
 [/INST] 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 $a^2+b^2\geq ab$.
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