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.
You don't require Cauchy Schwarz you can do it simply by observing how x and y varies given that it sums up to k