Problem: Let ( \ k) be a fixed odd positive integer.Find the minimum value of ( \ x^2+y^2),where ( \ x,y) are non-negative integers and ( \ x+y=k).

Solution: According to Cauchy Schwarz’s inequality,

we can write, ( \ (x^2+y^2)\times(1^2+1^2) \ge)(\ (x\times1+y\times1)^2)

=>( \ 2(x^2+y^2)\ge)(\ (x+y)^2)

=>( \ x^2+y^2\ge) (\frac{k^2}{2})

Therefore,the minimum value of ( \ x^2+y^2) is (\frac{k^2}{2}).

But it is given that (\ k) is a odd positive integer and (\ x,y \ge 0) so minimum value of  ( \ x^2+y^2) must be (\frac{k^2+1}{2}).

Concepts used:-Cauchy Schwarz’s inequality.