Prove the inequality (|x|+|y|)/(sqrt2)≤ |z| , where z=x+iy
Prove the inequality $\dfrac{|x|+|y|}{\sqrt2}\leq |z|$, where $z=x+iy$.
Explanation of Method
Square both sides (valid since both are non-negative) and reduce the inequality to $2|x||y|\leq x^2+y^2$, which is just $(|x|-|y|)^2\geq 0$ — always true. Equality holds exactly when $|x|=|y|$. The same result also follows directly from the Cauchy–Schwarz inequality applied to the vectors $(|x|,|y|)$ and $(1,1)$.
Method 1 · Direct algebraic proof
Step 1 · Write out $|z|$
$$ |z|=\sqrt{x^2+y^2} $$
We must show:
$$ \frac{|x|+|y|}{\sqrt2}\leq\sqrt{x^2+y^2} $$
Step 2 · Square both sides
Both sides are non-negative, so squaring preserves the inequality:
$$ \frac{(|x|+|y|)^2}{2}\leq x^2+y^2 $$
Step 3 · Expand and simplify
Since $|x|^2=x^2$ and $|y|^2=y^2$:
$$ \frac{x^2+2|x||y|+y^2}{2}\leq x^2+y^2 \quad\Longrightarrow\quad x^2+2|x||y|+y^2\leq 2x^2+2y^2 \quad\Longrightarrow\quad 2|x||y|\leq x^2+y^2 $$
Step 4 · Recognize a perfect square
$$ x^2+y^2-2|x||y|\geq 0 \quad\Longleftrightarrow\quad (|x|-|y|)^2\geq 0 $$
This is always true, so the chain of equivalent steps proves the inequality, with equality exactly when $|x|=|y|$.
$$ \boxed{\frac{|x|+|y|}{\sqrt2}\leq |z|} $$
Method 2 · Cauchy–Schwarz inequality
For $\mathbb{R}^2$, Cauchy–Schwarz states $(a_1b_1+a_2b_2)^2\leq(a_1^2+a_2^2)(b_1^2+b_2^2)$. Take $a_1=|x|$, $a_2=|y|$, $b_1=b_2=1$:
$$ (|x|+|y|)^2\leq(x^2+y^2)\cdot 2 \quad\Longrightarrow\quad |x|+|y|\leq\sqrt2\,|z| \quad\Longrightarrow\quad \frac{|x|+|y|}{\sqrt2}\leq|z| $$
Step 5 · Equality condition
Equality holds precisely when $|x|=|y|$, i.e. when $z$ lies on the lines $y=\pm x$ — in polar form $z=re^{i\theta}$, this is $|\cos\theta|=|\sin\theta|$, so:
$$ \theta=\frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4},\ \frac{7\pi}{4} $$
Manhattan path (green, length $|x|+|y|$) vs. straight-line Euclidean distance (blue, length $|z|$); dashed lines mark $y=\pm x$ where $\frac{|x|+|y|}{\sqrt2}=|z|$.
Geometric interpretation
$|x|+|y|$ is the Manhattan ($L^1$) distance from the origin to $(x,y)$; $\sqrt{x^2+y^2}$ is the Euclidean ($L^2$) distance. The inequality says the Manhattan distance is at most $\sqrt2$ times the Euclidean distance:
$$ \frac{\text{Manhattan distance}}{\sqrt2}\leq\text{Euclidean distance} $$
Like walking city blocks versus going "as the crow flies," the grid-path length is never more than $\sqrt2$ times the straight-line length, with the maximum ratio $\sqrt2$ attained exactly on the diagonals $y=\pm x$.
Verification with examples
| $z$ | $\frac{|x|+|y|}{\sqrt2}$ | $|z|$ | Comparison |
|---|---|---|---|
| $3+3i$ | $3\sqrt2\approx4.243$ | $\sqrt{18}=3\sqrt2\approx4.243$ | Equal ($|x|=|y|$) ✓ |
| $4+2i$ | $3\sqrt2\approx4.243$ | $\sqrt{20}=2\sqrt5\approx4.472$ | $4.243<4.472$ ✓ |
| $1+0i$ | $\frac{1}{\sqrt2}\approx0.707$ | $1$ | $0.707<1$ ✓ |
Result
Equality holds if and only if $|x|=|y|$, i.e. when $z$ lies on the lines making $45^{\circ}$ angles with the axes. Proof in brief: square both sides, reduce to $2|x||y|\leq x^2+y^2$, and recognize this as $(|x|-|y|)^2\geq0$.