**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.*

*Related*

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