Dots, ka \(x, y, z\) - pozitīvi skaitļi un \(xy+yz+zx>x+y+z\). Pierādīt, ka \(x+y+z>3\)
Ievērosim, ka
\((x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2xz+2yz=\)
\(=\frac{1}{2}\left[(x-y)^{2}+(x-z)^{2}+(y-z)^{2}\right]+3(xy+xz+yz) \geq 3(xy+xz+yz)>3(x+y+z)\). No \((x+y+z)^{2}>3(x+y+z)\) seko \(x+y+z>3\), k. b.j.