Question

cho x>=1/2,y>=1/2 cm x^2+y^2>=1/2

Answer 1

\(x^2>=\dfrac{1}{4}\)

\(y^2>=\dfrac{1}{4}\)

Do đó: \(x^2+y^2>=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

Answer 2

\(x\ge\dfrac{1}{2};y\ge\dfrac{1}{2}\)=>\(xy\ge\dfrac{1}{4}\)=>\(2xy\ge\dfrac{1}{2}\).

\(x+y\ge\dfrac{1}{2}+\dfrac{1}{2}=1\)

=>\(\left(x+y\right)^2\ge1\)

=>\(x^2+2xy+y^2\ge1\)

=>\(x^2+y^2\ge1-2xy\ge1-\dfrac{1}{2}=\dfrac{1}{2}\)

Answer 3

Ta có \(x\ge\dfrac{1}{2},y\ge\dfrac{1}{2}\)

=>\(xy\ge\dfrac{1}{4}\)

Ta có :\(\left(x-y\right)^2\ge0\)

=>\(x^2-2xy+y^2\ge0\)

=>\(x^2-2\dfrac{1}{4}+y^2\ge0\)

=>\(x^2+y^2\ge\dfrac{1}{2}\)