Question

[Cao Văn Hoang]

  Tìm A min. 

Answer 1

[Trần Tuấn Hoàng]

\(A=\dfrac{x}{\sqrt{y^2+1}}+\dfrac{y}{\sqrt{z^2+1}}+\dfrac{z}{\sqrt{x^2+1}}\)

\(=\dfrac{x}{\sqrt{y^2+xy+yz+zx}}+\dfrac{y}{\sqrt{z^2+xy+yz+zx}}+\dfrac{z}{\sqrt{x^2+xy+yz+zx}}\)

\(=\dfrac{x}{\sqrt{y\left(y+x\right)+z\left(y+x\right)}}+\dfrac{y}{\sqrt{z\left(z+x\right)+y\left(z+x\right)}}+\dfrac{z}{\sqrt{x\left(x+y\right)+z\left(x+y\right)}}\)

\(=\dfrac{x}{\sqrt{\left(y+x\right)\left(y+z\right)}}+\dfrac{y}{\sqrt{\left(z+x\right)\left(z+y\right)}}+\dfrac{z}{\sqrt{\left(x+y\right)\left(x+z\right)}}\)

\(\ge^{Caushy}\dfrac{x}{\dfrac{\left(y+x\right)+\left(y+z\right)}{2}}+\dfrac{y}{\dfrac{\left(z+x\right)+\left(z+y\right)}{2}}+\dfrac{z}{\dfrac{\left(x+y\right)+\left(x+z\right)}{2}}\)

\(=\dfrac{2x}{2y+x+z}+\dfrac{2y}{2z+x+y}+\dfrac{2z}{2x+y+z}\)

\(=2\left(\dfrac{x^2}{2yx+x^2+zx}+\dfrac{y^2}{2zy+xy+y^2}+\dfrac{z^2}{2xz+yz+z^2}\right)\)

\(\ge^{Caushy-Schwarz}2.\dfrac{\left(x+y+z\right)^2}{2yx+x^2+zx+2zy+xy+y^2+2xz+yz+z^2}\)

\(=2.\dfrac{\left(x+y+z\right)^2}{\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(xy+yz+zx\right)}\)

\(=2.\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+1}\)

Đặt \(\left(x+y+z\right)^2=t^2\). Ta có:

\(A\ge\dfrac{2t^2}{t^2+1}=\dfrac{2t^2+2-2}{t^2+1}=2-\dfrac{2}{t^2+1}\).

Ta lại có: \(t^2=\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=3.1=3\)

\(\Rightarrow A\ge2-\dfrac{2}{3+1}=\dfrac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{3}}{3}\)

Vậy \(MinA=\dfrac{3}{2}\)