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Sum of squares is nonnegative.

Prove that a^2 + b^2 \geq 0 with equality if and only if a = b = 0.


Proof. First, by Theorem I.20 we know that for any x \in \mathbb{R} we have if x \neq 0, then x^2 > 0. Further, since 0 \cdot 0 = 0, we have that if x = 0, then x^2 = 0. Thus, a^2 \geq 0 and b^2 \geq 0 and so a^2 + b^2 \geq 0.
Now, if a = b = 0, then a^2 + b^2 = 0 + 0 = 0. Conversely, if a^2 + b^2 = 0, then if a^2 > 0 we must have b^2 < 0 (otherwise the sum would be greater than 0). However, we know b^2 \geq 0, so this is a contradiction. Therefore, a^2 = 0, and hence, a^2 + b^2 = 0 \implies b^2 = 0, as well. By Theorem I.20 again, this must mean a = b =0 (since if a \neq 0 or b \neq 0 then a^2 or b^2 is greater than 0).
Hence, we indeed have a^2 + b^2 \geq 0 with equality if and only if a = b = 0. \qquad \blacksquare

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