Question

Cho x,y là 2 số thực dương. CMR:

\(\frac{x\sqrt{y}+y\sqrt{x}}{x+y}-\frac{x+y}{2}\le\frac{1}{4}\)

Answer 1

\(\frac{x\sqrt{y}+y\sqrt{x}}{x+y}-\frac{x+y}{2}\le\frac{x\sqrt{y}+y\sqrt{x}}{2\sqrt{xy}}-\frac{x+y}{2}=\frac{\sqrt{x}+\sqrt{y}}{2}-\frac{x+y}{2}\)

Cần chứng minh : \(\frac{\sqrt{x}+\sqrt{y}}{2}-\frac{x+y}{2}\le\frac{1}{4}\Leftrightarrow\sqrt{x}+\sqrt{y}-x-y\le\frac{1}{2}\Leftrightarrow2\sqrt{x}+2\sqrt{y}-2x-2y\le1\)

\(\Leftrightarrow2x+2y-2\sqrt{x}-2\sqrt{y}+1\ge0\)\(\Leftrightarrow\left(\sqrt{2x}-\frac{1}{\sqrt{2}}\right)^2+\left(\sqrt{2y}-\frac{1}{\sqrt{2}}\right)^2\ge0\) 

Vì BĐT cuối luôn đúng nên BĐT cần chứng minh luôn đúng khi x = y = \(\frac{1}{4}\)

Answer 2

\(VT=\frac{x\sqrt{y}+y\sqrt{x}}{x+y}-\frac{x+y}{2}\le\frac{\sqrt{2xy\left(x+y\right)}}{x+y}-\frac{x+y}{2}\)

\(\le\frac{\left(x+y\right)\sqrt{\frac{x+y}{2}}}{x+y}-\frac{x+y}{2}\) . Cm : \(\sqrt{\frac{x+y}{2}}-\frac{x+y}{2}\le\frac{1}{4}\)

Đặt \(x+y=t>0\)thì :

\(\sqrt{\frac{t}{2}}-\frac{t}{2}\le\frac{1}{4}\Leftrightarrow-\frac{1}{4}\left(\sqrt{2t}-1\right)^2\le0\) ( đúng )

Chúc bạn học tốt !!!