(A) dots, ka \(x+y=n\). Pierādīt, ka \(x^{2}+y^{2} \geq \frac{n^{2}}{2}\).
(B) dots, ka \(x+y+z=n\). Pierādīt, ka \(x^{2}+y^{2}+z^{2} \geq \frac{n^{2}}{3}\).
(A) \(x^{2}+y^{2}=x^{2}+(n-x)^{2}=2x^{2}-2nx+n^{2}=2\left(x-\frac{n}{2}\right)^{2}+\frac{n^{2}}{2} \geq \frac{n^{2}}{2}\), jo kvadrāts nav negatīvs.
(B) apzīmējam \(x=\frac{n}{3}+a, y=\frac{n}{3}+b, z=\frac{n}{3}+c\); tad \(a+b+c=(x+y+z)-3 \cdot \frac{n}{3}=0\). Tāpēc \(x^{2}+y^{2}+z^{2}=\left(\frac{n}{3}+a\right)^{2}+\left(\frac{n}{3}+b\right)^{2}+\left(\frac{n}{3}+c\right)^{2}=\frac{n^{2}}{3}+\frac{2}{3}n(a+b+c)+\left(a^{2}+b^{2}+c^{2}\right)=\frac{n^{2}}{3}+\left(a^{2}+b^{2}+c^{2}\right) \geq \frac{n^{2}}{3}\), k.b.j.