Minimum value of reciprocal squares I am bit stuck at a question. The question is : given: $x + y = 1$,  $x$ and $y$ both are positive numbers. What will be the minimum value of: $$\left(x + \frac{1}{x}\right)^2 + \left(y+\frac{1}{y}\right) ^2$$ 
I know placing $x = y$  will give the right solution. Is there any other solution?

$(x+1+y/x)^2+ (y+1+x/y)^2 >= 1/2 (x+1+y/x+y+1+x/y)^2  >= 1/2(2+1+x/y+y/x)^2 >=25/2 $
Equality holds when $x/y=y/x$ or $x=y$
