Trying to solve \begin{align} &W:= \left(x + \dfrac{1}{x}\right)^2 + \left(y + \dfrac{1}{y}\right)^2 \to \inf \\ &x+y = 1, \\ &x, y > 0. \end{align} Since $x, y \ne 0$, we can't use AM-GM(it will provide to solution with $x = 1$ and $y = 0$).
But we can use Cauchy-Schwarz inequality: $$\dfrac{1}{2}(1 + 1)\cdot\left( \left(x + \dfrac{1}{x}\right)^2 + \left(y + \dfrac{1}{y}\right)^2\right) \ge \dfrac{1}{2}\left( x + y + \dfrac{1}{x} + \dfrac{1}{y} \right)^2 = \dfrac{1}{2}\left( 1 + \dfrac{1}{x(1 -x)} \right)^2$$
So, since we want to find minimum, wee need $\dfrac{1}{x(1 -x)}$ to be also minumum, i.e. $(x (1 - x)) $ got to be maximum.
Maximum for $(x (1 - x))$ is achieved only when $x= 0.5$, so infimum for W is $12.5$ with $x = y = 0.5$.
Am i right here?
Yes, this is correct. Since $W$ attains the minimum value of $25/2$ for the choice $x = y = 1/2$, we know the lower bound is tight.