Mk viết thiếu đề :
Cho các số thực x , y thỏa mãn :
x2 + 3xy + 4y2 \(\le\dfrac{7}{2}\)
Chứng minh rằng x + y \(\le2\)
Đặt x + y = t suy ra x = t - y
Do đó x2 + 3xy + 4y2 \(\le\)\(\dfrac{7}{2}\)
\(\Leftrightarrow\left(t-y\right)^2+3\left(t-y\right)y+4y^2-\dfrac{7}{2}\le0\)
\(\Leftrightarrow2y^2+ty+t^2-\dfrac{7}{2}\le0\Leftrightarrow\left(2y+\dfrac{t}{2}\right)^2\le7\left(1-\dfrac{t^2}{4}\right)\)
Suy ra \(1-\dfrac{t^2}{4}\ge0\) ( vì \(\left(2y+\dfrac{t}{2}\right)^2\ge0\) )
\(\Rightarrow t^2\le4\Rightarrow x+y\le2\) . Dấu bằng xảy ra khi
\(\left\{{}\begin{matrix}x+y=2\\\left(2y+\dfrac{t}{2}\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+y=2\\\left(2y+\dfrac{x+y}{2}\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
