Áp dụng bất đẳng thức Cô-si, ta có:
\(P=5x+3y+\dfrac{10}{x}+\dfrac{8}{y}\)
\(=\left(\dfrac{5x}{2}+\dfrac{10}{x}\right)+\left(\dfrac{3y}{6}+\dfrac{8}{y}\right)+\left(\dfrac{5x}{2}+\dfrac{15y}{6}\right)\)
\(=\left(\dfrac{5x}{2}+\dfrac{10}{x}\right)+\left(\dfrac{3y}{6}+\dfrac{8}{y}\right)+\dfrac{15\left(x+y\right)}{6}\)
\(\ge2.\sqrt{\dfrac{5x}{2}.\dfrac{10}{x}}+2.\sqrt{\dfrac{3y}{6}.\dfrac{8}{y}}+\dfrac{15.6}{6}\)
\(=2\sqrt{25}+2\sqrt{4}+15\)
\(=2.5+2.2+15\)
\(=10+4+15\)
\(=29\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5x}{2}=\dfrac{10}{x}\\\dfrac{3y}{6}=\dfrac{8}{y}\\x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Vậy \(MinP=29\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Lời giải:
Áp dụng BĐT Cô-si:
$\frac{y}{2}+\frac{8}{y}\geq 2\sqrt{\frac{y}{2}.\frac{8}{y}}=4$
$\frac{5}{2}x+\frac{10}{x}\geq 2\sqrt{\frac{5x}{2}.\frac{10}{x}}=10$
$\frac{5}{2}(x+y)\geq \frac{5}{2}.6=15$
Cộng theo vế các BĐT trên thu được:
$P\geq 29$
Vậy $P_{\min}=29$ khi $(x,y)=(2,4)$
