Với \(x>0;x\ne1\):
\(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
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Khi đó: \(P+6\sqrt{x}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}+6\sqrt{x}\)
\(=2\sqrt{x}+2+\dfrac{2}{\sqrt{x}}+6\sqrt{x}\)
\(=\left(8\sqrt{x}+\dfrac{2}{\sqrt{x}}\right)+2\)
Áp dụng BĐT Cauchy cho hai số dương \(8\sqrt{x}\) và \(\dfrac{2}{\sqrt{x}}\), ta được:
\(8\sqrt{x}+\dfrac{2}{\sqrt{x}}\ge2\sqrt{8\sqrt{x}\cdot\dfrac{2}{\sqrt{x}}}=8\)
\(\Rightarrow\left(8\sqrt{x}+\dfrac{2}{\sqrt{x}}\right)+2\ge8+2=10\)
\(\Rightarrow P+6\sqrt{x}\ge10\)
Dấu \("="\) xảy ra khi: \(8\sqrt{x}=\dfrac{2}{\sqrt{x}}\Leftrightarrow8x=2\Leftrightarrow x=\dfrac{1}{4}\) (tmđk)
Vậy \(P+6\sqrt{x}_{min}=10\) tại \(x=\dfrac{1}{4}\).
