ĐKXĐ:\(y\ge0\)
Ta có: \(\dfrac{y+4}{\sqrt{y}+2}=\dfrac{\left(\sqrt{y}-2\right)\left(\sqrt{y}+2\right)+8}{\sqrt{y}+2}=\sqrt{y}+2+\dfrac{8}{\sqrt{y}+2}\)
\(=\sqrt{y}+2+\dfrac{8}{\sqrt{y}+2}-4\)
Áp dụng BĐT Cô-si cho 2 số nguyên dương ta dc:
\(\sqrt{y}+2+\dfrac{8}{\sqrt{y}+2}\ge4\sqrt{2}\)
\(\Leftrightarrow\sqrt{y}+2+\dfrac{8}{\sqrt{y}+2}-4\ge4\sqrt{2}-4\)
Dấu "=" xảy ra khi \(y=12-8\sqrt{2}\)
\(P=\dfrac{y-4+8}{\sqrt{y}+2}\)
\(P=\dfrac{\left(\sqrt{y}+2\right)\left(\sqrt{y}-2\right)+8}{\sqrt{y}+2}\)
\(P=\sqrt{y}-2+\dfrac{8}{\sqrt{y}+2}\)
\(P=\sqrt{y}+2-4+\dfrac{8}{\sqrt{y}+2}\)
vì \(\sqrt{y}+2\ge0\)
\(\dfrac{8}{\sqrt{y}+2}\ge0\)
Áp dụng bđt Cauchy cho 2 số dương \(\sqrt{y}+2\) và \(\dfrac{8}{\sqrt{y}+2}\), ta có:
\(\sqrt{y}+2+\dfrac{8}{\sqrt{y}+2}\ge2\sqrt{\left(\sqrt{y}+2\right).\dfrac{8}{\sqrt{y}+2}}\)
\(\sqrt{y}+2+\dfrac{8}{\sqrt{y}+2}\ge2.2\sqrt{2}\)
\(\sqrt{y}+2+\dfrac{8}{\sqrt{y}+2}\ge4\sqrt{2}\)
\(\sqrt{y}+2+\dfrac{8}{\sqrt{y}+2}-4\ge4\sqrt{2}-4\)
\(P\ge4\sqrt{2}-4\)
Dấu '=' xảy ra \(\Leftrightarrow\sqrt{y}+2=\dfrac{8}{\sqrt{y}+2}\)
\(\Leftrightarrow\left(\sqrt{y}+2\right)^2=8\)
\(\Leftrightarrow\sqrt{y}+2=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{y}=2\sqrt{2}-2\)
\(\Leftrightarrow y=12-8\sqrt{2}\)