Question

Cho \(a,b,c\in\left[0,1\right]\) và \(a+b+c=2\)

Tìm max \(A=a^4+b^4+c^4+\dfrac{11}{2}abc\)

Answer 1

Xét \(0 \geq (a-1)(2a-1)^2(a+2) = 4a^4 - 11a^2 + 9a - 2\)

\(\Rightarrow 4\left(a^4+b^4+c^4\right) \leq 11\left(a^2 + b^2 + c^2\right) - 9(a+b+c) + 6\)

\(\Rightarrow a^4+b^4+c^4\leq \dfrac{11}{4}\left(a^2+b^2+c^2\right) - 3\).

Ngoài ra, ta có \((a-1)(b-1)(c-1)\leq 0\Rightarrow abc \leq ab+bc+ca - 1\).

Do đó \(P\le\dfrac{11}{4}\left(a^2+b^2+c^2+2ab+2bc+2ca-2\right)-3=\dfrac{5}{2}\).

Đẳng thức xảy ra khi \((a,b,c) =\left(0, \dfrac{1}{2}. \dfrac{1}{2}\right)\) và các hoán vị.