Question

Cho x,y,z là các số dương và x+y+z \(\le\)1.Chứng minh:

Answer 1

Áp dụng BĐT Mincopxki và AM-GM ta có:

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

\(\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{9}{x+y+z}\right)^2}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\dfrac{81}{\left(x+y+z\right)^2}}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\dfrac{1}{\left(x+y+z\right)^2}+\dfrac{80}{\left(x+y+z\right)^2}}\)

\(\ge\sqrt{2\sqrt{\left(x+y+z\right)^2\cdot\dfrac{1}{\left(x+y+z\right)^2}}+80}\)

\(\ge\sqrt{2+80}=\sqrt{82}\)

Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{3}\)