P = x2 + 5y2 + 2xy – 4x – 8y + 2020 tìm gtnn

$P = x^2+5y^2+2xy-4x-8y+2020$

$ = (x^2+2xy+y^2) – 4.(x+y) + 4y^2-4y+2020$

$ = (x+y)^2-2.(x+y).2+2^2 + 4y^2-4y+2016$

$ = (x+y-2)^2 + (4y^2-4y+1)+2015$

$ = (x+y-2)^2+(2y-1)^2+2015 $

Vì $(x+y-2)^2 ≥0 ∀ x,y$ ; $(2y-1)^2 ≥0 ∀y$

$⇒ (x+y-2)^2+(2y-1)^2 ≥ 0 ∀x,y$

$⇒(x+y-2)^2+(2y-1)^2+2015 ≥ 2015 ∀x,y$

Dấu “=” xảy ra $⇔\left\{ \begin{array}{l}x+y-2=0\\2y-1=0\end{array} \right.$

$⇔\left\{ \begin{array}{l}x+\dfrac{1}{2}=2\\y=\dfrac{1}{2}\end{array} \right.$ $⇔\left\{ \begin{array}{l}x=\dfrac{3}{2}\\y=\dfrac{1}{2}\end{array} \right.$

Vậy $P_{min} = 2015$ tại $x=\dfrac{3}{2}$ và $y=\dfrac{1}{2}$