violympic vog 17 kho

V

vipboycodon

Ta có: $x+4y = 2$ => $x = 2-4y$
* $x^2+4y^2 $
= $4-16y+16y^2+4y^2$
= $20y^2-16y+4$
= $20(y^2-\dfrac{4y}{5}+\dfrac{1}{5})$
= $20(y^2-2y.\dfrac{2}{5}+\dfrac{4}{25}+\dfrac{1}{25})$
= $20(y-\dfrac{2}{5})^2+\dfrac{4}{5} \ge \dfrac{4}{5}$
Dấu "=" xảy ra khi $x = y = \dfrac{2}{5}$
 
Last edited by a moderator:
E

eye_smile

Cách khác:
AD Cauchy-Schwarz, có:
$4={(x+2.2y)^2}$ \leq $(1+4)({x^2}+4{y^2})=5({x^2}+4{y^2})$
\Rightarrow ${x^2}+4{y^2}$ \geq $\dfrac{4}{5}$
Dấu "=" xảy ra \Leftrightarrow $x=y=\dfrac{2}{5}$
 
Top Bottom