Câu hỏi của Lunox Butterfly Seraphim

Question

Tìm a,b,c nguyên biết: $a^2+b^2+c^2+4\le ab+3b+2c$

Diệu Huyền · Accepted Answer

Ta có: $a^2+b^2+c^2+4\le ab+3b+2c$

Hay: $\left(a^2-ab+\frac{b^2}{4}\right)+\left(\frac{3b^2}{4}-3b+3\right)+\left(c^2-2c+1\right)\le0$

$\Leftrightarrow\left(a-\frac{b}{2}\right)^2+3\left(\frac{b}{2}-1\right)^2+\left(c-1\right)^2\le0$

$\Leftrightarrow\left\{{}\begin{matrix}c-1=0\\frac{b}{2}-1=0\a-\frac{b}{2}=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}a=1\b=2\c=1\end{matrix}\right.$

Vậy .....................Câu trả lời: Ta có: $a^2+b^2+c^2+4\le ab+3b+2c$

Hay: $\left(a^2-ab+\frac{b^2}{4}\right)+\left(\frac{3b^2}{4}-3b+3\right)+\left(c^2-2c+1\right)\le0$

$\Leftrightarrow\left(a-\frac{b}{2}\right)^2+3\left(\frac{b}{2}-1\right)^2+\left(c-1\right)^2\le0$

$\Leftrightarrow\left\{{}\begin{matrix}c-1=0\\frac{b}{2}-1=0\a-\frac{b}{2}=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}a=1\b=2\c=1\end{matrix}\right.$

Vậy .....................