Ta có :(a+b+c)2(a+b+c)2-3(ab+bc+ca)≥≥0
<=> a22+b22+c22+2ab+2bc+2ca-3ab-3bc-3ca≥≥0
<=>1212(2a22+2b22+2c22-2ab-2bc-2ca)≥≥0
<=>1212[(a2a2-2ab+b22)+(b22-bc+c22)+(c22-2ca+a22)]≥≥0
<=> 1212[(a-b)22+(b-c)22+(c-a)22]≥≥0
<=>(a+b+c)22≥≥3(ab+bc+ca)
Xét hiệu : \(\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\)
\(=a^2+b^2+c^2+2ab+2bc+2ca-3ab-3bc-3ca\)
\(=a^2+b^2+c^2-ab-bc-ca\)
\(=\dfrac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)
\(=\dfrac{1}{2}\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ca+c^2\right)\right]\)
=\(\dfrac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\ge0\)
Vậy: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Ta có :\(\left(a+b+c\right)^2\)-3(ab+bc+ca)\(\ge\)0
<=> a\(^2\)+b\(^2\)+c\(^2\)+2ab+2bc+2ca-3ab-3bc-3ca\(\ge\)0
<=>\(\dfrac{1}{2}\)(2a\(^2\)+2b\(^2\)+2c\(^2\)-2ab-2bc-2ca)\(\ge\)0
<=>\(\dfrac{1}{2}\)[(\(a^2\)-2ab+b\(^2\))+(b\(^2\)-bc+c\(^2\))+(c\(^2\)-2ca+a\(^2\))]\(\ge\)0
<=> \(\dfrac{1}{2}\)[(a-b)\(^2\)+(b-c)\(^2\)+(c-a)\(^2\)]\(\ge\)0
<=>(a+b+c)\(^2\)\(\ge\)3(ab+bc+ca)
