# non-calculus proof that $|x| + |y| \le \sqrt{2} |z|$

I would like to see a nice, non-calculus proof that $|x| + |y| \le \sqrt{2} |z|$, where $z$ is the complex number $x + iy$. The more elementary the better, avoiding even trig if possible. Thank you.

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How about formulating it as $$\frac{|x|+|y|}{2} \leq \sqrt{\frac{|x|^2+|y|^2}{2}}$$ and recognizing it as Cauchy–Schwarz in different garb? – J. M. Oct 19 '11 at 2:50

Let $a, b \in \mathbb{R}$. Your inequality follows from:

\begin{align} (|a|-|b|)^2 \geq 0 &\Longrightarrow a^2 + b^2 \geq 2|a||b| \\ &\Longrightarrow 2(a^2 + b^2) \geq (|a|+|b|)^2 \\ &\Longrightarrow \sqrt{2} \sqrt{a^2 + b^2} \geq |a| + |b| \end{align}

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(This is pretty similar to DJC's answer)

Squares of real numbers are non-negative:

\begin{align} ( |x| + |y| )^2 & \leq ( |x|+|y| )^2 + (|x|-|y|)^2 \\ & = (|x|^2 + |y|^2 + 2|xy|) + (|x|^2 + |y|^2 - 2|xy|) \\ &= 2|x|^2 + 2|y|^2 \\ &= 2|z|^2 \end{align}

Taking square roots:

$$|x|+|y| \leq \sqrt{2}|z|$$

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By dividing by $|z|$, it suffices to prove: For any unit modulus complex number, $x + iy$, we have $|x| + |y| \leq \sqrt{2}$. A unit modulus complex number is represented by a point $(x, y)$ on the unit circle. The maximum of the sum of $|x|$ and $|y|$ must occur when $|x|=|y|$ (why?), at which point Pythagoras tells us that $|x|=|y|= \sqrt{2}/2$, and hence $|x|+|y| = \sqrt{2}$.

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