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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.