Question

1. Cho \(x,y,z>0\)\(xyz=1\)

Tìm min P= \(x+y+z+\frac{13}{3\left(xy+yz+zx\right)}\)

2. Cho \(a>0\)

Tìm min P= \(\frac{a^4+a^3+3a^2+a+1}{a^3+a}\)

Answer 1

a/ \(P\ge x+y+z+\frac{13}{\left(x+y+z\right)^2}=\frac{13\left(x+y+z\right)}{27}+\frac{13\left(x+y+z\right)}{27}+\frac{13}{\left(x+y+z\right)^2}+\frac{1}{27}\left(x+y+z\right)\)

\(P\ge3\sqrt[3]{\frac{13^3\left(x+y+z\right)^2}{27^2\left(x+y+z\right)^2}}+\frac{1}{27}.3\sqrt[3]{xyz}=\frac{40}{9}\)

\(P_{min}=\frac{40}{9}\) khi \(x=y=z=1\)

2/Chia cả tử và mẫu cho \(a^2\):

\(P=\frac{a^2+\frac{1}{a^2}+2+a+\frac{1}{a}+1}{a+\frac{1}{a}}=\frac{\left(a+\frac{1}{a}\right)^2+a+\frac{1}{a}+1}{a+\frac{1}{a}}\)

Đặt \(a+\frac{1}{a}=x\ge2\)

\(\Rightarrow P=\frac{x^2+x+1}{x}=x+\frac{1}{x}+1=\frac{x}{4}+\frac{1}{x}+\frac{3x}{4}+1\)

\(\Rightarrow P\ge2\sqrt{\frac{x}{4x}}+\frac{3.2}{4}+1=\frac{7}{2}\)

\(P_{min}=\frac{7}{2}\) khi \(x=2\) hay \(a=1\)