Question

cho 3 số thực dương a b c thỏa mãn \(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\) . tìm giá trị nhỏ nhất của K= \(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\)

Answer 1

\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Rightarrow\frac{b}{a}+\frac{b}{c}=2\)

Đặt \(\left\{{}\begin{matrix}\frac{b}{a}=x\\\frac{b}{c}=y\end{matrix}\right.\) \(\Rightarrow x+y=2\) \(\Rightarrow\left\{{}\begin{matrix}x=2-y\\y=2-x\end{matrix}\right.\)

\(K=\frac{1+\frac{b}{a}}{2-\frac{b}{a}}+\frac{1+\frac{b}{c}}{2-\frac{b}{c}}=\frac{1+x}{2-x}+\frac{1+y}{2-y}=\frac{1+x}{y}+\frac{1+y}{x}\)

\(K=\frac{1}{x}+\frac{1}{y}+\frac{x}{y}+\frac{y}{x}\ge\frac{4}{x+y}+2\sqrt{\frac{xy}{yx}}=2+2=4\)

\(\Rightarrow K_{min}=4\) khi \(x=y=1\) hay \(a=b=c\)