Question

Cho x,y>0 ; x+y<=6

Tìm minB=\(\dfrac{x^2y+xy^2+24x+6y}{xy}\).

Answer 1

\(B=x+y+\dfrac{6}{x}+\dfrac{24}{y}=\left(\dfrac{3x}{2}+\dfrac{6}{x}\right)+\left(\dfrac{3y}{2}+\dfrac{24}{y}\right)-\dfrac{3}{2}\left(x+y\right)\)

\(B\ge2\sqrt{\dfrac{18x}{2x}}+2\sqrt{\dfrac{72y}{2y}}-\dfrac{3}{2}.6=15\)

\(B_{min}=15\) khi \(\left(x;y\right)=\left(2;4\right)\)