Prove that |Re(a)|+|Im(a)|≤|a|sqrt2 for every complex number a
Prove that $\ |\text{Re}(a)|+|\text{Im}(a)|\le|a|\sqrt2\ $ for every complex number $a$.
Explanation of Method
Write $a=x+iy$ and apply the Cauchy–Schwarz inequality to the real vectors $(|x|,|y|)$ and $(1,1)$; equivalently, verify directly that $2|xy|\le x^2+y^2$, which is just $(|x|-|y|)^2\ge0$.
Step 1 · Reduce to a real inequality
Let $a=x+iy$ with $x,y\in\mathbb R$. The claim $|x|+|y|\le\sqrt2\sqrt{x^2+y^2}$ has both sides nonnegative, so it is equivalent (after squaring) to:
$$ (|x|+|y|)^2\le2(x^2+y^2) \quad\Longleftrightarrow\quad x^2+2|xy|+y^2\le2x^2+2y^2 \quad\Longleftrightarrow\quad 2|xy|\le x^2+y^2 $$
Step 2 · Prove $2|xy|\le x^2+y^2$
This is equivalent to $(|x|-|y|)^2\ge0$, expanding to $x^2-2|x||y|+y^2\ge0$, i.e. $x^2+y^2\ge2|xy|$ — always true since a square is nonnegative.
Step 3 · Alternative proof via Cauchy–Schwarz
Apply Cauchy–Schwarz to the vectors $(|x|,|y|)$ and $(1,1)$:
$$ (|x|\cdot1+|y|\cdot1)^2\le(x^2+y^2)(1^2+1^2)=2(x^2+y^2) $$
Taking square roots recovers Step 1 directly.
Step 4 · Equality condition
Equality in $(|x|-|y|)^2\ge0$ holds exactly when $|x|=|y|$, i.e. when $a$ makes an angle of $\pm45^\circ$ or $\pm135^\circ$ with the real axis.
Verification
| $a$ | $|\text{Re}(a)|+|\text{Im}(a)|$ | $|a|\sqrt2$ |
|---|---|---|
| $1+i$ | $2$ | $\sqrt2\cdot\sqrt2=2$ (equality) |
| $1$ | $1$ | $\sqrt2\approx1.414$ |
| $3+4i$ | $7$ | $5\sqrt2\approx7.07$ |
Result
The "taxicab" sum of the components is at most $\sqrt2$ times the Euclidean modulus, tight exactly on the diagonals — a direct instance of the equivalence of the $\ell^1$ and $\ell^2$ norms on $\mathbb R^2$.