Cho Q= $\frac{(x+27)P}{(\sqrt{x}+3)(\sqrt{x}-2)}$, với P= $\sqrt{x}-2$. Đk:$\left \{ {{x\geq0} \atop {x\neq1, x\neq4 }} \right.$. CM Q$\geq$ 6

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1. `Q=((x+27)P)/((sqrtx+3)(sqrtx-2))(x>=0,x ne 1,x ne 4)`

   `=((x+27)(sqrtx-2))/((sqrtx+3)(sqrtx-2))`

   `=(x+27)/(sqrtx+3)`

   `=(x-9+36)/(sqrtx+3)`

   `=((sqrtx+3)(sqrtx-3)+36)/(sqrtx+3)`

   `=sqrtx-3+36/(sqrtx+3)`

   `=sqrtx+3+36/(sqrtx+3)-6`

   Áp dụng BĐT cosi với 2 số dương ta có:

   `(sqrtx+3)+36/(sqrtx+3)>=2.6=12`

   `=>Q>=12-6=6(đpcm)`

   Dấu “=” xảy ra khi `(sqrtx+3)=36/(sqrtx+3)`

   `<=>(sqrtx+3)^2=36`

   `<=>sqrtx+3=6`(do `sqrtx+3>=3`)

   `<=>sqrtx=3`

   `<=>x=9`

   Vậy `Q>=6(AA x>=0,x ne 1,x ne 4)`

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