Question

Cho \(4x+y=1\)

Chung minh \(4x^2+y^2\ge\frac{1}{5}\)

Answer 1

         Áp dụng BĐT Bunhia-copxki:

Ta có: (4x+y)²=(2x.2+y.1)²\(\le\)(4x²+y²)(2²+1²)

    <=> 1\(\le\)(4x²+y²).5

       => 4x²+y² \(\ge\frac{1}{5}\)(đpcm)

Answer 2

\(\text{Ta có : }\)

\(4x+1=1< 4x^2+y^2\)

\(\text{Mà }1>\frac{1}{5}=0,2\)

\(\Rightarrow\text{ }4x^2+y^2>\frac{1}{5}\)