Constrained minimum with inequalities Show that for all real positive $x$ and $y$ such that $x+y=1$, the following holds: $$\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\geq \frac{25}{2}.$$
I know this can probably be easily solved with Lagrange multipliers, but I am more interested in proving it using inequalities. I think it could be profitable to use AM-HM or something similar but I am not sure about how to proceed.
How can I do it?
Thanks.

This problem can be rewritten as:
$$\sqrt{\frac{(x+\frac{1}{x})^2+(y+\frac{1}{y})^2}{2}} \geq\frac{5}{2}$$
We can use inequality between square and arithmetic mean:
$$\sqrt{\frac{(x+\frac{1}{x})^2+(y+\frac{1}{y})^2}{2}} \geq \frac{(x+\frac{1}{x})+(y+\frac{1}{y})}{2}=\frac{1+\frac{1}{x}+\frac{1}{y}}{2}$$
We only have to show, that $$\frac{1}{x}+\frac{1}{y} \geq 4$$
But this comes from $x+y=1$ and inequality between arithmetic and harmonic mean:
$$\frac{1}{2}=\frac{x+y}{2}\geq\frac{2}{\frac{1}{x}+\frac{1}{y}} \Rightarrow \frac{1}{x}+\frac{1}{y} \geq 4$$
