\(4x^2+y^2=4x^2+\left(1-4x\right)^2=4x^2+1-8x+16x^2=20x^2-8x+1=20\left(x^2-\frac{2}{5}x+\frac{1}{20}\right)\)
\(=20\left[x^2-\frac{2}{5}x+\frac{1}{25}+\frac{1}{100}\right]=20\left(x-\frac{1}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{5}\)
BĐT$\Leftrightarrow 20x^2+5y^2\geq (4x+y)^2=16x^2+8xy+y^2\Leftrightarrow 2(x-y)^2\geq 0$ (đúng)
Dấu "=" xảy ra khi $x=y=\frac{1}{5}$
4x + y = 1 => y = 1 - 4x
Nên : 4x^2 + y^2 = 4x^2 + (1 - 4x)^2
= 4x^2 + 1 - 8x + 16x^2
= 20x^2 - 8x + 1
= 4(5x^2 - 2x) + 1
= 4/5(25x^2 - 10x) + 1
= 4/5(25x^2 - 2.5x + 1) + 1/5
= 4/5(5x - 1)^2 + 1/5
>= 1/5
Dấu "=" xảy ra khi x = 1/5 => y = 1/5
Áp dụng BĐT Bunyakovsky cho 2 bộ số (2x,y) và (2,1) ta có:
\(\left(4x^2+y^2\right)\left(2^2+1\right)=\left(\left(2x\right)^2+y^2\right)\left(2^2+1\right)\)
\(\ge\left(\left(2x\cdot2\right)+y\cdot1\right)^2=\left(4x+1\right)^2=1^2=1\)
\(\Leftrightarrow\left(4x^2+y^2\right)\cdot5\ge1\)
\(\Leftrightarrow4x^2+y^2\ge\dfrac{1}{5}\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=\dfrac{1}{5}\)
Vậy \(4x^2+y^2\ge\dfrac{1}{5}\)