ĐKXĐ: \(x>0\)
\(S=\dfrac{x+\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}}\)
\(=\left(\sqrt{x}+\dfrac{4}{\sqrt{x}}\right)+1\)
Áp dụng BĐT Cauchy - Schwarz cho hai số dương \(\sqrt{x}\) và \(4\sqrt{x}\), ta được:
\(\sqrt{x}+\dfrac{4}{\sqrt{x}}\ge2\sqrt{\sqrt{x}\cdot\dfrac{4}{\sqrt{x}}}=4\)
\(\Rightarrow\left(\sqrt{x}+\dfrac{4}{\sqrt{x}}\right)+1\ge4+1=5\)
\(\Rightarrow S\ge5\)
Dấu "=" xảy ra khi: \(\sqrt{x}=\dfrac{4}{\sqrt{x}}\Leftrightarrow x=4\) (tmđk)
Vậy \(S_{min}=5\) tại \(x=4\).
ĐK: x > 0
\(S=\dfrac{x+\sqrt{x}+4}{\sqrt{x}}=\dfrac{x}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}}+\dfrac{4}{\sqrt{x}}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}}\)
Vì x > 0\(\Rightarrow\sqrt{x}>0\Rightarrow\dfrac{4}{\sqrt{x}}>0\)
Áp dụng BĐT Cô-si ta có:
\(\sqrt{x}+\dfrac{4}{\sqrt{x}}\ge2\sqrt{\sqrt{x}\cdot\dfrac{4}{\sqrt{x}}}=2\sqrt{4}=4\)
\(\Rightarrow\sqrt{x}+\dfrac{4}{\sqrt{x}}+1\ge4+1=5\)
Dấu ''='' xảy ra\(\Leftrightarrow\sqrt{x}=\dfrac{4}{\sqrt{x}}\Leftrightarrow x=4\)
Vậy Min S = 5 \(\Leftrightarrow\) x = 4
