Find all triples $(\alpha, \beta, \gamma)$ of positive real numbers for which the expression $$K = \frac{\alpha+3 \gamma}{\alpha + 2\beta + \gamma} + \frac{4\beta}{\alpha+\beta+2\gamma} - \frac{8 \gamma}{\alpha+ \beta + 3\gamma}$$obtains its minimum value.
Problem
Source: Cyprus IMO TST 2018, Problem 3
Tags: inequalities
01.07.2018 16:43
Amir Hossein wrote: Find all triples $(\alpha, \beta, \gamma)$ of positive real numbers for which the expression $$K = \frac{\alpha+3 \gamma}{\alpha + 2\beta + \gamma} + \frac{4\beta}{\alpha+\beta+2\gamma} - \frac{8 \gamma}{\alpha+ \beta + 3\gamma}$$obtains its minimum value. Solution of Zhangyanzong: $$K =2\left( \frac{\alpha+\beta+2\gamma}{\alpha + 2\beta + \gamma}+ \frac{2(\alpha + 2\beta + \gamma)}{\alpha+\beta+2\gamma}\right) +4\left( \frac{\alpha+ \beta + 3\gamma}{\alpha+\beta+2\gamma}+ \frac{2(\alpha+\beta+2\gamma)}{\alpha + \beta + 3\gamma}\right) -17\geq12\sqrt{2}-17.$$......
01.07.2018 16:45
Amir Hossein wrote: Find all triples $(\alpha, \beta, \gamma)$ of positive real numbers for which the expression $$K = \frac{\alpha+3 \gamma}{\alpha + 2\beta + \gamma} + \frac{4\beta}{\alpha+\beta+2\gamma} - \frac{8 \gamma}{\alpha+ \beta + 3\gamma}$$obtains its minimum value. China girls MO 2004.