If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that $(a+1/a)^2+(b+1/b)^2\ge 25/2$ 
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that 
  $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$  

My work:
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}$$
Now, we have $a^2+\dfrac{1}{a^2}\ge 2$ and $b^2+\dfrac{1}{b^2}\ge 2$.
Here, I am stuck, I cannot use the information provided, $a+b=1$ to any use. Please help!

By QM-AM, $\displaystyle a^2+b^2 \geq \frac{1}{2}(a+b)^2 = \frac{1}{2}$.  QM-AM again gives $\displaystyle \frac{1}{a^2}+\frac{1}{b^2} \geq \frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)^2$.  By AM-HM, $\displaystyle \frac{2}{\frac{1}{a}+\frac{1}{b}} \leq \frac{a+b}{2} = \frac{1}{2}$, whence $\displaystyle\frac{1}{a}+\frac{1}{b} \geq 4$, and thus $\displaystyle \frac{1}{a^2}+\frac{1}{b^2} \geq 8$.  Thus we have $$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 = a^2 + b^2 + \frac{1}{a^2}+\frac{1}{b^2} + 4 \geq \frac{1}{2} + 8 + 4 = \frac{25}{2}.$$
