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.

yes,use $Cuachy-Schwarz$ inequality,we have 
$$\left[\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2\right][1+1]\ge\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2=\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2$$
use $Cauchy-Schwarz $ inequality,we have
$$\left(\dfrac{1}{a}+\dfrac{1}{b}\right)(a+b)\ge(1+1)^2$$
