Question

Cho các số dương a, b, c thỏa mãn \(a^2+b^2+c^2+abc=4\). CM\(a+b+c\le3\)

Answer 1

TK: \(a^2+b^2+c^2+abc=4\Leftrightarrow a^2+abc+b^2+c^2-4=0\)

Coi đây là PT bậc 2 ẩn a

\(\Leftrightarrow\Delta=b^2c^2-4b^2-4c^2+16=\left(4-b^2\right)\left(4-c^2\right)\\ \Leftrightarrow a=\dfrac{-bc+\sqrt{\left(4-b^2\right)\left(4-c^2\right)}}{2}\le\dfrac{-bc+\dfrac{4-b^2+4-c^2}{2}}{2}=\dfrac{8-\left(b+c\right)^2}{4}\\ \Leftrightarrow a+b+c\le\dfrac{8-\left(b+c\right)^2}{4}+b+c=\dfrac{8-\left(b+c\right)^2+4\left(b+c\right)}{4}=\dfrac{12-\left(b+c-2\right)^2}{4}\le3\)