Áp dụng bđt Cauchy:
$\sqrt[]{x}\leq \dfrac{x+1}{2}$
$\sqrt[]{y-1}\leq \dfrac{y-1+1}{2}=\dfrac{y}{2}$
$\sqrt[]{z-2}\leq \dfrac{z-2+1}{2}=\dfrac{z-1}{2}$
Do đó: $\sqrt[]{x}+\sqrt[]{y-1}+\sqrt[]{z-2}\leq \dfrac{x+1+y+z-1}{2}=\dfrac{x+y+z}{2}$
Dấu "=" xảy ra khi và chỉ khi: $x=1;y=2;z=3$
Vậy: $x^2+y^2+z^2=1^2+2^2+3^2=14$