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 $a = b = 0. \qquad \blacksquare$

Stumbling Robot

Sum of squares is nonnegative.

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