Question

Tìm giá trị nhỏ nhất của biểu thức \(Q=\frac{a^2}{b^2+\left(a+b\right)^2}+\frac{b^2}{a^2+\left(a+b\right)^2}\)

Answer 1

@Nguyễn Việt Lâm

Answer 2

ĐK: \(a;b\) ko đồng thời bằng 0

\(Q\ge\frac{a^2}{b^2+2\left(a^2+b^2\right)}+\frac{b^2}{a^2+2\left(a^2+b^2\right)}=\frac{a^2}{2a^2+3b^2}+\frac{b^2}{3a^2+2b^2}\)

Đặt \(\left\{{}\begin{matrix}2a^2+3b^2=x>0\\3a^2+2b^2=y>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=\frac{3y-2x}{5}\\b^2=\frac{3x-2y}{5}\end{matrix}\right.\)

\(\Rightarrow Q\ge\frac{3y-2x}{5x}+\frac{3x-2y}{5y}=\frac{3}{5}\left(\frac{y}{x}+\frac{x}{y}\right)-\frac{4}{5}\ge\frac{3}{5}.2\sqrt{\frac{xy}{xy}}-\frac{4}{5}=\frac{2}{5}\)

\(\Rightarrow Q_{min}=\frac{2}{5}\) khi \(x=y\) hay \(a=b\)