Prove that $\frac{1}{xy}\ge4$ given that $x+y=1$ and conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge2$ 
Let $x,y\in\mathbb R^+$ and $x+y=1$
1- Prove that $\frac{1}{xy}\ge4$
2- Conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge25$

I have tried to start from $x+y=1$ or $x\ge0\land y\ge0$ and reach $\frac{1}{xy}\ge4$ but with no result.
Update:
I've proved the first part of the question (Thanks to Jack's comment).
Since the second question says "Conclude" that means I have to use the first proof. I tried to square the first proof and got it close to the second question, but
again no result. There is probably a trick that I don't know.

\begin{eqnarray*}
x+y &=& 1 \\
x^2+2xy+y^2  &=& 1 \\
x^2-2xy+y^2  &=& 1-4xy \\
0 \leq (x-y)^2  &=&  1 -4xy.
\end{eqnarray*}
Now rearrange and we have the desired inequality.
For the second part 
\begin{eqnarray*}
\left(\frac{1}{x} -\frac{1}{y}\right)^2 \geq 0 \\
\frac{1}{x^2} +\frac{1}{y^2} \geq \frac{2}{xy}
\end{eqnarray*}
So 
\begin{eqnarray*}
1+ \frac{1}{x^2} +\frac{1}{y^2} + \left(\frac{1}{xy}\right)^2 \geq 1+ \frac{2}{xy}  + \left(\frac{1}{xy}\right)^2 =\left(1+\frac{1}{xy}\right)^2 \geq 25.\\
\end{eqnarray*}
