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How to prove that $$a^2+b^2 \geqslant 2ab$$ is true, where $a$ and $b$ are both real numbers?

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marked as duplicate by Martin Sleziak, Micah, Mike Miller, Davide Giraudo, Alex S Jul 6 at 20:33

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In this question of yours I showed you why that is true. Maybe the only detail missing is that for any $x\in\Bbb R$, $x^2\geq 0$. –  Pedro Tamaroff Nov 21 '12 at 4:02
    
    
I think its the other way around. –  wythagoras Jul 6 at 15:00

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up vote 5 down vote accepted

Remember that for any $x \in \mathbb{R}$, we have that $x^2 \geq 0$ i.e. square of any real number is non-negative.

Hence, $$(a-b)^2 \geq 0$$ since $a,b \in \mathbb{R}$. Expand $(a-b)^2$ and rearrange to get what you want. \begin{align} (a-b)^2 & \geq 0\\ a^2 + b^2 - 2ab & \geq 0\\ a^2 + b^2 & \geq 2ab \end{align}

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Remember that any real number squared is non-negative, and

$$a^2+b^2\geq 2ab\Longleftrightarrow a^2-2ab+b^2=(a-b)^2\geq 0\,\,...$$

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