Question

CMR:

a. (a+b+c)² >= 3(ab+bc+ca)²

b.(a+b+c)² =< 3(a²+b²+c²)

Answer 1

b.

\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

Xét hiệu:

\(3\left(a^2+b^2+c^2\right)-\left(a+b+c^2\right)=3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ac\)

\(=2a^2+2b^2+2c^2-2ab-2ac-2bc\)

\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)

\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\) ( luôn đúng)

=> \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

Answer 2