Library / /5 of 34

, 2010

Algebra Difficulty 4.9 AIME Prove it Austria

Let $x$ and $y$ be positive real numbers with $x + y = 1$ .
Prove that
$\frac{(3x - 1)^2}{x} + \frac{(3y - 1)^2}{y} \geq 1.$
When does equality hold?

Solution

We have
$\begin{aligned} \frac{(3x - 1)^2}{x} + \frac{(3y - 1)^2}{y} &= \frac{9x^2 - 6x + 1}{x} + \frac{9y^2 - 6y + 1}{y} \\ &= 9x - 6 + \frac{1}{x} + 9y - 6 + \frac{1}{y} \\ &= -3 + \frac{1}{x} + \frac{1}{y}. \end{aligned}$
It remains to show that
$\frac{1}{x} + \frac{1}{y} \geq 4.$
This is a consequence of the inequality between the arithmetic and the harmonic mean and the condition $x + y = 1$ :
$(x + y) \left( \frac{1}{x} + \frac{1}{y} \right) \geq 4.$
Equality holds exactly for $x = y$ and therefore for $x = y = 1/2$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.