Math inequality proof If $a, b$ are positive real numbers and $a + b = 1$, prove that
$$
\left(a +\frac{1}{a}\right)^2 + \left(b +\frac{1}{b}\right)^2 \geq \frac{25}{2}
$$
Thank you.

It's easily seen that, for $a+b=1$ with $a,b\gt0$, the expression
$$ab+{1\over ab}$$
is minimized when $a=b=1/2$.  (In general, $x+{1\over x}$ is smallest when $x$ is as close to $1$ as possible.  If $a+b=1$, the closest $ab$ gets to $1$ is when $a=b=1/2$.)  It follows that
$$\begin{align}
\left(a+{1\over a} \right)^2+\left(b+{1\over b} \right)^2&=a^2+b^2+{1\over a^2}+{1\over b^2}+4\cr
&=(a-b)^2+\left({1\over a}-{1\over b} \right)^2+2\left(ab+{1\over ab}\right)+4\cr
&\ge 2\left(ab+{1\over ab}\right)+4\cr
&\ge 2\left({1\over4}+{1\over1/4}\right)+4\cr
&={25\over2}
\end{align}$$
