# Why is this inequality true? $(a+b)^2\leq 2(a^2+b^2)$

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?

$$(a-b)^2\geq0\\a^2-2ab+b^2\geq0\\a^2+b^2\geq2ab\\2a^2+2b^2\geq a^2+2ab+b^2\\2(a^2+b^2)\geq(a+b)^2$$
• Thanks! But how did you go from $a^2+b^2\geq 2ab$ to $2a^2+2b^2\geq a^2+2ab+b^2$? – JDoeDoe Mar 2 '17 at 8:21
• @JDoeDoe He added $a^{2} + b^{2}$ to both sides. – Mattos Mar 2 '17 at 8:21
Equivalently: $$a^2+2ab+b^2 \leq 2a^2+2b^2$$ Equivalently: $$0 \leq a^2-2ab+b^2$$ Equivalently: $$0 \leq (a-b)^2$$
Because by C-S $$2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geq(1\cdot a+1\cdot b)^2=(a+b)^2$$