Ta có: a2+b2\(\ge2ab\)
b2+c2\(\ge2bc\)
a2+c2\(\ge2ac\)
=> \(2a^2+2b^2+2c^2\ge2ab+2bc+2ac\)
<=> \(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
<=> \(a^2+b^2+c^2\ge ab+bc+ac\)
Áp dụng BĐT cô si ta có:
\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\) (1)
\(b^2+c^2\ge2\sqrt{b^2c^2}=2bc\) (2)
\(c^2+a^2\ge2\sqrt{c^2a^2}=2ac\) (3)
(1)+(2)+(3) vế theo vế ta được:
\(2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ac\) (đpcm)