TITLE: Minimum value of reciprocal squares
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: $x+y=1,~$ x and y both are positive numbers.

Then $x=\sin^2t$ and $y=\cos^2t$ is a valid parameterization, wouldn't you agree ?


What will be the minimum value of $~\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2~$ ?

Hint: Rewrite the expression in question in terms of t, and ask yourself what techniques you already know for finding the extrema $($minima and maxima$)$ of functions of a single variable.