Question

[Phạm Khánh Hồng]

Cho a, b, c bất kỳ, hãy so sánh a^2 + b^2 + c^2 với ab + bc + ca

Answer 1

[Nobi Nobita]

Xét hiệu, ta có:

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

\(=\frac{1}{2}.\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)

\(=\frac{1}{2}.\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\right]\)

\(=\frac{1}{2}.\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

Vì \(\left(a-b\right)^2\ge0\); \(\left(b-c\right)^2\ge0\); \(\left(c-a\right)^2\ge0\)\(\forall a,b,c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

\(\Rightarrow\frac{1}{2}.\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\forall a,b,c\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)