(A) Dots, ka \(s+t=p\). Pierādīt, ka \(2s^{2} \geq p^{2}-2t^{2}\).
(B) Dots, ka \(s+t+u=p\). Pierādīt, ka \(3s^{2} \geq p^{2}-3t^{2}-3u^{2}\).
(A) \((s+t)^{2}=p^{2} \Rightarrow s^{2}+2st+t^{2}=p^{2} \Rightarrow s^{2}+2st-t^{2}=p^{2}-2t^{2} \Rightarrow 2s^{2}-(s-t)^{2}=p^{2}-2t^{2}\). Tā kā \((s-t)^{2} \geq 0\), tad \(2s^{2} \geq p^{2}-2t^{2}\), k.b.j.
(B) \((s+t+u)^{2}=p^{2} \Rightarrow s^{2}+t^{2}+u^{2}+2st+2su+2tu=p^{2} \Rightarrow\) \(s^{2}-2t^{2}-2u^{2}+2st+2su+2tu=p^{2}-3t^{2}-3u^{2} \Rightarrow 3s^{2}-(s-t)^{2}-(s-u)^{2}-(t-u)^{2}=p^{2}-3t^{2}-3u^{2}\).
Tā kā \((s-t)^{2} \geq 0,(s-u)^{2} \geq 0\) un \((t-u)^{2} \geq 0\), tad \(3s^{2} \geq p^{2}-3t^{2}-3u^{2}\), k.b.j.