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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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    [In this question of yours](http://math.stackexchange.com/questions/241717/prove-or-disprove-c-is-a-subset-of-d/241725#241725) I showed you why that is true. Maybe the only detail missing is that for any $x\in\Bbb R$, $x^2\geq 0$.2012-11-21
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    See also [Show that for all real numbers $a$ and $b$, $ab \le (1/2)(a^2+b^2)$](http://math.stackexchange.com/questions/943994/show-that-for-all-real-numbers-a-and-b-ab-le-1-2a2b2)2015-07-06
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    I think its the other way around.2015-07-06

2 Answers 2

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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

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