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, 2019

Algebra Difficulty 7.0 National olympiad, round 2 Prove it Greece

If $\alpha$ , $\beta$ , $\gamma$ are positive real numbers such that $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = 3$ , prove that
$\frac{\alpha + \beta}{\alpha^2 + \alpha\beta + \beta^2} + \frac{\beta + \gamma}{\beta^2 + \beta\gamma + \gamma^2} + \frac{\gamma + \alpha}{\gamma^2 + \gamma\alpha + \alpha^2} \le 2 .$
When equality is valid?

Solution

By putting $x = \frac{1}{\alpha}$ , $y = \frac{1}{\beta}$ , $z = \frac{1}{\gamma}$ , we have $x + y + z = 3$ and the inequality takes the form:
$\frac{xy(x+y)}{x^2+xy+y^2} + \frac{yz(y+z)}{y^2+yz+z^2} + \frac{zx(z+x)}{z^2+zx+x^2} \le 2.$
We have $\frac{xy}{x^2+xy+y^2} \le \frac{1}{3} \Leftrightarrow x^2+xy+y^2 \ge 3xy \Leftrightarrow (x-y)^2 \ge 0$ and equality is valid when $x=y$ . Multiplying both sides by $x+y>0$ , we get
$\frac{xy(x+y)}{x^2+xy+y^2} \le \frac{1}{3}(x+y). \qquad (1)$
Similarly we get the relations:
$\frac{yz(y+z)}{y^2+yz+z^2} \le \frac{1}{3}(y+z) \quad (2), \quad \frac{zx(z+x)}{z^2+zx+x^2} \le \frac{1}{3}(z+x). \quad (3)$
Finally using summation of (1), (2) and (3) we have
$\frac{xy(x+y)}{x^2+xy+y^2} + \frac{yz(y+z)}{y^2+yz+z^2} + \frac{zx(z+x)}{z^2+zx+x^2} \le \frac{1}{3} \cdot 2(x+y+z) = 2.$
Equality is valid when $x = y = z = 1$ or $\alpha = \beta = \gamma = 1$ .

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