Cho $a,b,c$ là các số thực không âm thỏa mãn :$ 0 \leq a;b;c \leq 2$ và $a+b+c=3$. Tìm Max:$A=a^3+b^3+c^3$
Không mất tính tổng quát, giả sử $2 \geq a \geq b \geq c \geq 0$ Đặt $a = 1 + x$; $c= 1 - y$; do đó, $b=1+y-x$ Đồng thời, ta cũng thu được, $0 \leq x ; y \leq 1$ Cần phải chứng minh: Có, $A=a^3+b+3+c^3$ Hay, $A=(1+x)^3+(1-y)^3+(1+y-x)^3$ $=x^3+3x^2+3x+1+1-3y+3y^2-y^3+1+3.(y-x)+3.(y-x)^2+(y-x)^3$ $=x^3+3x^2+3x+1+1-3y+3y^2-y^3+1+3y-3x+3y^2-6xy+3y^2+y^3-3y^2x+3yx^2-x^3$ $=6x^2+6y^2-6xy+3-3y^2x+3yx^2$ Mà $x,y\epsilon[0;1]$ nên $x^2 \leq x;y^2 \leq y$ $A=3.(2x^2+2y^2-2xy+1-xy(y-x))$ Xét $B=2x^2+2y^2-2xy+1-xy(y-x)$ $B-3 \leq 2x^2+2y^2-4xy-xy(y-x)=(y-x)(2y-2x-xy)$ Dễ dàng cm $B-3\leq0$ nên, $A\leq 9$ Dấu bằng xảy ra khi và chỉ khi $y=x=1$ hay $a=2;b=1;c=0$

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