\(P=5x+3y+\dfrac{12}{x}+\dfrac{16}{y}\)
\(P=3x+\dfrac{12}{x+y}+\dfrac{16}{y}+2.\left(x+y\right)\)
Áp dụng BĐT Cauchy ta có:
\(3x+\dfrac{12}{x}\ge2\sqrt{\left(3.12\right)}=12\)
\(y+\dfrac{16}{y}\ge8\)
Lại có: \(2\left(x+y\right)\ge2.6=12\)
\(\Rightarrow P\ge12+8+12=32\)
Dấu " = " xảy ra \(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{12}{x}\\y=\dfrac{16}{y}\\x+y=6\end{matrix}\right.\)
\(\Rightarrow x=2;y=4\)
Vậy \(P_{Min}=32\Leftrightarrow\left[{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
P=\(5x+3y+\dfrac{12}{x}+\dfrac{16}{y}\)
=\(3x+\dfrac{12}{x}+y+\dfrac{16}{y}+2\left(x+y\right)\)
AD BĐT cô si :
Ta có \(3x+\dfrac{12}{x}\ge2\sqrt{3x.\dfrac{12}{x}}=2\sqrt{36}=12\)
\(y+\dfrac{16}{y}\ge2\sqrt{y.\dfrac{16}{y}}=2\sqrt{16}=8\)
\(2\left(x+y\right)\ge2.6=12\)
=> P\(\ge12+8+12=32\)
Dấu = xra \(\left\{{}\begin{matrix}3x=\dfrac{12}{x}\\y=\dfrac{16}{y}\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\left(x;y\right)=\left(2;4\right)\)
Vậy GTNN của P=32 khi (x;y)=(2;4)