Question

giải bất phương trình: \(\dfrac{\left(x+y\right)^2}{4}+\dfrac{x+y}{2}\ge a\sqrt{b}+b\sqrt{a}\)

Answer 1

Viết nhầm biến thôi :v. Sửa''ss

\(\dfrac{\left(x+y\right)^2}{4}+\dfrac{x+y}{2}\ge x\sqrt{y}+y\sqrt{x}\)

Ta có: \(VT\ge\dfrac{4xy}{4}+\dfrac{x}{2}+\dfrac{x}{2}=\dfrac{xy}{2}+\dfrac{xy}{2}+\dfrac{x}{2}+\dfrac{y}{2}\)

\(\Leftrightarrow VT\ge2\sqrt{\dfrac{xy}{2}.\dfrac{x}{2}}+2\sqrt{\dfrac{xy}{2}.\dfrac{y}{2}}\)

\(\Leftrightarrow VT\ge x\sqrt{y}+y\sqrt{x}\)(đpcm)

Đẳng thức xảy ra khi \(\left[{}\begin{matrix}x=y=1\\x=y=0\end{matrix}\right.\)

Answer 2

sai đề rồi