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.

given $X + Y = 1$, proof that $\frac{1}{XY} \ge 4$, where $X$ and $Y$ are real positive integers.
Since $X$ and $Y$ are positive integers then, $X \lt 1$ and $Y \lt 1$ for they to sum to $1$.
$X$ and $Y$ can be both equal 
So that $X + Y = 1$
this then becomes $X$ or $Y$ $= \frac{1}{2} = 0.5$
But if $X$ and $Y$ are distinct,
then
$
\begin{align}
X \gt 0.5 & and & Y \lt   0.5 & or\\
X \lt 0.5  &  and & Y \gt 0.5
\end{align}
$
Just depending of which is greater and lesser, for convenience let's assume $Y \gt X$.
So that a number $" a "$ must lie between both $X$ and $Y$ 
Such that....
$X = 0.5 - a$ and $Y = 0.5 + a$
where
$$
0 \le a \lt \frac{1}{2}
$$
because $X$ and $Y$ are positive integers.      So that the product of $X$ and  $Y$ is more pronounced as
$(0.5-a)×(0.5+a)$
which gives
$(0.5^2 - a^2)$
Therefore the inverse of there product
$\frac{1}{XY}$ becomes $\frac{1}{0.5^2 - a^2}$
Since $a \lt \frac{1}{2}$, $a$ runs from $\frac{1}{2}$ to approach $0$
         $\frac{1}{\frac{1}{4} - a^2}$
its maximum value is when $a$ approach $0$.                      $\frac{1}{0.25 - a^2}$
$\frac{1}{XY} \ge \frac{1}{0.25}$                                          $\frac{1}{XY} \ge 4$
