(A) Dots, ka \(a+b=c\). Pierādīt, ka \(2a^{2} \geqq c^{2}-2b^{2}\).
(B) Dots, ka \(a+b+c=d\). Pierādīt, ka \(3a^{2} \geq d^{2}-3b^{2}-3c^{2}\).
(A) \((a+b)^{2}=c^{2} \Rightarrow a^{2}+2ab+b^{2}=c^{2} \Rightarrow a^{2}+2ab-b^{2}=c^{2}-2 b^{2} \Rightarrow\) \(2 a^{2}-(a-b)^{2}=c^{2}-2b^{2}\)
Tā kā \((a-b)^{2} \geq 0\), tad \(2a^{2} \geq c^{2}-2b^{2}\), k.b.j.
(B) \((a+b+c)^{2}=d^{2} \Rightarrow a^{2}+b^{2}+c^{2}+2ab+2ac+2bc=d^{2} \Rightarrow\) \(a^{2}-2b^{2}-2c^{2}+2ab+2ac+2bc=d^{2}-3b^{2}-3c^{2} \Rightarrow\) \(3a^{2}-(a-b)^{2}-(a-c)^{2}-(b-c)^{2}=d^{2}-3b^{2}-3c^{2}\).
Tā kā \((a-b)^{2} \geq 0,(a-c)^{2} \geq 0\) un \((b-c)^{2} \geq 0\), tad \(3a^{2} \geq d^{2}-3b^{2}-3c^{2}\), k.b.j.