Question

Cho a,b,c≥0 thỏa mãn a+b+c=1. Tìm Max \(N=ab+3ac+5bc\)

Answer 1

Lời giải:

$N=a(b+3c)+5bc=(1-b-c)(b+3c)+5bc$

$=b+3c-b^2-3c^2+bc$

$-N=b^2+3c^2-bc-b-3c$

$-2N=2b^2+6c^2-2bc-2b-6c$

$\geq b^2+5c^2-2b-6c$

$=(b+c-1)^2+(2c-1)^2-2bc-2$

$\geq -2(bc+1)$

Mà $bc\leq \frac{(b+c)^2}{4}\leq \frac{1}{4}$

$\Rightarrow bc+1\leq \frac{5}{4}$

$\Rightarrow -2(bc+1)\geq \frac{-10}{4}$
$\Rightarrow -2N\geq \frac{-10}{4}$

$\Rightarrow N\leq \frac{5}{4}$

Vậy $N_{\max}=\frac{5}{4}$ khi $(a,b,c)=(0,\frac{1}{2}, \frac{1}{2})$