Question

Tìm min của biểu thức sau \(x+y+\dfrac{6}{x}+\dfrac{24}{y}\) biết x,y>0;x+y<=6

Answer 1

Ta có: \(P+\frac{1}{2}(a+b)=(\frac{3}{2}x+\frac{6}{x})+(\frac{3}{2}y+\frac{24}{y})\geq 2.3+2.6=18\)

Mà \(a+b\leq 6\) suy ra \(P\geq 15\)

dấu = xảy ra

\(<=> x=2 , y=4\)

Answer 2

Đặt A = ( \(\dfrac{3x}{2}\) + \(\dfrac{6}{x}\) ) + ( \(\dfrac{3y}{2}\) + \(\dfrac{24}{y}\) ) - ( \(\dfrac{x+y}{2}\) )

Áp dụng BĐT Cô-si ta có

\(\dfrac{3x}{2}+\dfrac{6}{x}\ge6\)

\(\dfrac{3y}{2}+\dfrac{24}{y}\ge6\)

Có x + y \(\le6\)

=> - (x + y) \(\ge6\) => \(\dfrac{-\left(x+y\right)}{2}\ge3\)

=> A \(\ge15\)

Dấu " = " xảy ra <=> x = 2; y = 4