Question

Cho a + b + c = 3. Cmr : \(a^2+b^2+c^2\ge3\)

Answer 1

Với mọi số thực a;b;c ta luôn có:

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

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)

\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}.3^2=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)