Question

Cho hai số dương a,b thỏa mãn \(a+\frac{1}{b}=1\). Cmr:

\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2}\)

Answer 1

Đặt \(\left\{{}\begin{matrix}a=x\\\frac{1}{b}=y\end{matrix}\right.\) \(\Rightarrow x+y=1\)

\(VT=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2\)

\(VT\ge\frac{1}{2}\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+y+\frac{4}{x+y}\right)^2=\frac{1}{2}\left(1+4\right)^2=\frac{25}{2}\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\) hay \(\left\{{}\begin{matrix}a=\frac{1}{2}\\b=2\end{matrix}\right.\)