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Why is this inequality true? $a,b$ are real numbers. $$ (a+b)^2=a^2+2ab+b^2\leq 2(a^2+b^2) $$ I know $(a+b)^2=a^2+2ab+b^2 \geq 0$, but then?

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$$(a-b)^2\geq0\\a^2-2ab+b^2\geq0\\a^2+b^2\geq2ab\\2a^2+2b^2\geq a^2+2ab+b^2\\2(a^2+b^2)\geq(a+b)^2$$

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  • $\begingroup$ Thanks! But how did you go from $a^2+b^2\geq 2ab$ to $2a^2+2b^2\geq a^2+2ab+b^2$? $\endgroup$ – JDoeDoe Mar 2 '17 at 8:21
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    $\begingroup$ @JDoeDoe He added $a^{2} + b^{2}$ to both sides. $\endgroup$ – Mattos Mar 2 '17 at 8:21
  • $\begingroup$ @JDoeDoe Exactly. $\endgroup$ – 76david76 Mar 2 '17 at 8:22
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Equivalently: $$a^2+2ab+b^2 \leq 2a^2+2b^2$$ Equivalently: $$0 \leq a^2-2ab+b^2$$ Equivalently: $$0 \leq (a-b)^2$$

This latter expression is true: squares are nonnegative.

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Because by C-S $$2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geq(1\cdot a+1\cdot b)^2=(a+b)^2$$

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