Tìm max $A=a^2+b^2+c^2$, biết $\left\{\begin{matrix}-1\leq a,b,c\leq 3 & \\ a+b+c=1 & \end{matrix}\right.$
Edited by hoanglong2k, 28-10-2015 - 17:20.
Tìm max $A=a^2+b^2+c^2$, biết $\left\{\begin{matrix}-1\leq a,b,c\leq 3 & \\ a+b+c=1 & \end{matrix}\right.$
Edited by hoanglong2k, 28-10-2015 - 17:20.
$-1\leq a\leq 3\Rightarrow (a+1)(a-3)\leq 0\Leftrightarrow a^{2}-2a-3\leq 0$
Tương tự ta có $b^{2}-2b-3\leq 0$ $c^{2}-2c-3\leqslant 0$
Cộng vế theo vế ta được $a^{2}+b^{2}+c^{2}-2(a+b+c)-9\leq 0\Rightarrow a^{2}+b^{2}+c^{2}\leqslant 9+2.1=11$
Vậy max A=11 <=>$\left\{\begin{matrix} a=-1 & & \\ b=-1 & & \\ c=3& & \end{matrix}\right.$(giả sử $c\geq a\geq b$)
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