Olympiad Maths Prep

Track / Stage 6 / 56 of 400 #1056 of 2000

Problem 1056

National olympiad, first round
Algebra Difficulty 6.1 Prove it

Example 2.1.2. Suppose that x,y,z≥1x, y, z \geq 1 and 1x+1y+1z=2\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2. Prove that
x+y+z≥x−1+y−1+z−1\sqrt{x+y+z} \geq \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution. By hypothesis, we obtain
x−1x+y−1y+z−1z=1.\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1 .

According to Cauchy-Schwarz, we have
∑cycx=(∑cycx)(∑cycx−1x)≥(∑cycx−1)2\sum_{cyc} x=\left(\sum_{cyc} x\right)\left(\sum_{cyc} \frac{x-1}{x}\right) \geq\left(\sum_{cyc} \sqrt{x-1}\right)^{2}
which implies
x+y+z≥x−1+y−1+z−1\sqrt{x+y+z} \geq \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.