Question

[Phạm Thúy Vy]

Với mọi x, y, z >= 0 . Chứng minh rằng

\(\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{6\left(x+y+z\right)}\)

Answer 1

[Kuro Kazuya]

Áp dụng bất đẳng thức Mincopski

\(\Rightarrow\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{\left(x+y+z\right)^2+9}\)

Chứng minh rằng \(\sqrt{\left(x+y+z\right)^2+9}\ge\sqrt{6\left(x+y+z\right)}\)

\(\Leftrightarrow\left(x+y+z\right)^2+9\ge6\left(x+y+z\right)\)

\(\Leftrightarrow\dfrac{\left(x+y+z\right)^2+9}{x+y+z}\ge6\)

\(\Leftrightarrow x+y+z+\dfrac{9}{x+y+z}\ge6\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow x+y+z+\dfrac{9}{x+y+z}\ge2\sqrt{\dfrac{9\left(x+y+z\right)}{x+y+z}}=2\sqrt{9}=6\) ( đpcm )

Vậy \(\sqrt{\left(x+y+z\right)^2+9}\ge\sqrt{6\left(x+y+z\right)}\)

Mà \(\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{\left(x+y+z\right)^2+9}\)

\(\Rightarrow\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{6\left(x+y+z\right)}\) ( đpcm )

Dấu " = " xảy ra khi \(x=y=z=1\)