Question

Cho x > 4 ; x ≥ 0 . Tìm GTNN của P = \(\dfrac{2x}{\sqrt{x}-2}\)

Answer 1

\(P=\dfrac{2x}{\sqrt{x}-2}=\dfrac{2x-8+8}{\sqrt{x}-2}=\dfrac{2\left(x-4\right)}{\sqrt{x}-2}+\dfrac{8}{\sqrt{x}-2}=2\left(\sqrt{x}+2\right)+\dfrac{8}{\sqrt{x}-2}=2\left(\sqrt{x}-2\right)+\dfrac{8}{\sqrt{x}-2}+8\)

Áp dụng BĐT Cauchy cho các số dương , ta có :
\(2\left(\sqrt{x}-2\right)+\dfrac{8}{\sqrt{x}-2}\ge2\sqrt{2\left(\sqrt{x}-2\right).\dfrac{8}{\sqrt{x}-2}}=2.\sqrt{16}=2.4=8\)

\(\Leftrightarrow2\left(\sqrt{x}-2\right)+\dfrac{8}{\sqrt{x}-2}+8\ge8+8=16\)

\(\Rightarrow P_{Min}=16."="\Leftrightarrow x=16\left(TM\right)\)