Question

Cho a;b;c thỏa mãn: \(a+b+c=\frac{3}{2}\)

Chứng minh rằng: \(a^2+b^2+c^2\ge\frac{3}{4}\)

Answer 1

 Bạn thử chứng minh kiểu này đi :

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

Answer 2

Mình chứng minh theo cách trên :3

\(\frac{a^2+b^2+c^2}{3}-\left(\frac{a+b+c}{3}\right)^2=\frac{a^2+b^2+c^2}{3}-\frac{\left(a+b+c\right)^2}{9}\)

\(=\frac{1}{9}\left[3\left(a^2+b^2+c^2\right)-\left(a+b+c\right)^2\right]\)

\(=\frac{1}{9}\left[2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\right]\)

\(=\frac{1}{9}\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+c^2\right)\right]\)

\(=\frac{1}{9}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)