Question

Cho x+y+z=xyz. Tìm Min A= \(\frac{y}{x\sqrt{y^2+1}}+\frac{z}{y\sqrt{z^2+1}}+\frac{x}{z\sqrt{x^2+1}}\)

Answer 1

Ta có \(\frac{y}{x\sqrt{y^2+1}}=\frac{y\sqrt{xz}}{x\sqrt{y\left(x+y+z\right)+xz}}=\frac{yz}{\sqrt{x\left(y+z\right).z\left(x+y\right)}}\ge\frac{2yz}{2xz+xy+yz}\)

Đặt \(a=xy,b=yz,c=xz\)=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Khi đó

\(P\ge\frac{2b}{2c+a+b}+\frac{2c}{2a+b+c}+\frac{2a}{2b+a+c}\ge\frac{2\left(a+b+c\right)^2}{b^2+c^2+a^2+3\left(ab+bc+ac\right)}\)

Xét \(P\ge\frac{3}{2}\)

=> \(4\left(a+b+c\right)^2\ge3\left(a^2+b^2+c^2\right)+9\left(ab+bc+ac\right)\)

<=> \(a^2+b^2+c^2\ge\left(ab+bc+ac\right)\)(luôn đúng )

Vậy \(MinP=\frac{3}{2}\)khi a=b=c=3=> \(x=y=z=\sqrt{3}\)