Question

[Nguyễn Thiều Công Thành]

với a;b;c là các số thực dương thỏa mãn a+b+c=1.Tìm Max P=6(ab+bc+ca)+a(a-b)²+b(b-c)²+c(c-a)²

Answer 1

[LIVERPOOL]

P=\(\left(a^2+b^2+c^2+2ab+2ac+2bc\right)+4\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)\)\(+a^3+b^3+c^3-2\left(a^2b+b^2c+c^2a\right)+ab^2+bc^2+ca^2\)\(=1+4\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)\left(a+b+c\right)+\left(a^3+b^3+c^3\right)\)\(-2\left(a^2b+b^2c+c^2a\right)+\left(ab^2+bc^2+ca^2\right)\)\(=1+4\left(ab+bc+ca\right)-3\left(a^2b+b^2c+c^2a\right)\)

Mà \(\left(a^2b+b^2c+c^2a\right)\left(b+c+a\right)\ge\left(ab+bc+ca\right)^2\)

=> \(P\le1+4\left(ab+bc+ca\right)-3\left(ab+bc+ca\right)^2\). Đặt \(ab+bc+ca=t\le\frac{1}{3}\)

=> \(P\le-3\left(t^2-\frac{2}{3}t+\frac{1}{9}\right)+2t+\frac{4}{3}\le-3\left(t-\frac{1}{3}\right)^2+\frac{2}{3}+\frac{4}{3}\le2\)

Dấu bằng xảy ra khi \(t=\frac{1}{3}\)<=> \(a=b=c=\frac{1}{3}\)