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Irish Mathematical Olympiad

Algebra Difficulty 4.3 AIME Prove it Ireland

Prove for all complex numbers zz that
∣z∣2+2∣z−1∣≥1, |z|^2 + 2|z - 1| \ge 1,
with equality iff z=1z = 1.

Solution

If ∣z∣≥1|z| \ge 1, then
∣z∣2+2∣z−1∣≥1+2∣z−1∣≥1, |z|^2 + 2|z - 1| \ge 1 + 2|z - 1| \ge 1,
with equality iff z=1z = 1.

If ∣z∣<1|z| < 1, then
∣z−1∣≥∣z∣−1=1−∣z∣, |z - 1| \ge |z| - 1 = 1 - |z|,
and so
∣z∣2+2∣z−1∣−1≥∣z∣2+2(1−∣z∣)−1=(∣z∣−1)2>0. |z|^2 + 2|z - 1| - 1 \ge |z|^2 + 2(1 - |z|) - 1 = (|z| - 1)^2 > 0.

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