Áp dụng BĐT bunhiacopxki
\(\left(1+2^2\right)\left(x^2+4y^4\right)\ge\left(x+4y\right)^2\)
<=> \(5\left(x^2+4y^2\right)\ge1\)
<=> \(x^2+4y^2\ge\dfrac{1}{5}\) (đpcm)
dấu '=' xảy ra khi x=\(\dfrac{y}{4}\) => x=\(\dfrac{13}{17}\) ;y=\(\dfrac{4}{17}\)
Áp dụng BĐT bunhiacopxki ta có
\(\left(1^2+2^2\right)\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)
<=> \(5\left(x^2+4y^2\right)\ge1\) (vì x+4y=1)
<=> \(x^2+4y^2\ge\dfrac{1}{5}\) (đpcm)
dấu "=" xảy ra khi x=\(\dfrac{y}{4}\) => \(x-\dfrac{y}{4}=0\) (1)
ta có x+4y=1 (2)
(2) - (1) ta đc
\(x+4y-\left(x-\dfrac{y}{4}\right)=1\)
<=>\(x+4y-x+\dfrac{y}{4}=1\)
<=> \(\dfrac{16y}{4}+\dfrac{y}{4}=\dfrac{4}{4}\)
<=> 16y+y=4
<=> 17y=4
<=> y=\(\dfrac{4}{17}\)
=> x=\(\dfrac{13}{15}\)
