Question

[pham linh chi]

tìm min P= 4( x^2 + y^2) + 1/x+y với x,y>0, x+y>=2

Answer 1

[Đoàn Đức Hà]

\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(P=4\left(x^2+y^2\right)+\frac{1}{x+y}\ge2\left(x+y\right)^2+\frac{1}{x+y}\)

\(=\frac{31}{16}\left(x+y\right)^2+\frac{1}{16}\left(x+y\right)^2+\frac{1}{2\left(x+y\right)}+\frac{1}{2\left(x+y\right)}\)

\(\ge\frac{31}{16}.2^2+3\sqrt[3]{\frac{1}{16}\left(x+y\right)^2.\frac{1}{2\left(x+y\right)}.\frac{1}{2\left(x+y\right)}}\)

\(=\frac{17}{2}\)

Dấu \(=\)khi \(x=y=1\).