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.
 A: \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*}
A: Given that $x+y=1$ means simply that the combination $(x,y)$ has to be a point on that line. Not above, not below, but ON the line.
With this condition the easiest thing to do is to graph: The ONLY point that is on both the line and on $\frac{1}{xy}=4$ is the point $(1/2,1/2)$ and this is the only solution that works. Perhaps you meant to ask "Solve $\frac{1}{xy}=4$,given that $x+y\geq1$?. Then the whole branch in the first quadrant would work. But as of now, it is just one single point
A: 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 $0$.                      $\frac{1}{0.25 - a^2}$
$\frac{1}{XY} \ge \frac{1}{0.25}$                                          $\frac{1}{XY} \ge 4$
A: There's many answers here, but I'll post my answer here anyways; the proof essentially shows that even when some expression is the smallest it can be, the inequality still holds, which means it must necessarily always hold.
Given that $x+ y = 1$ and $x,y \in \mathbb{R}^+$, show that $\frac{1}{xy} \ge 4$.
The smallest value $\frac{1}{xy}$ can have is when we have the largest value $xy$ can have. Now what is the largest possible value for $xy$? Change the given assumption to $y = 1-x$ and substitute into $xy$ to get $x-x^2$. We know that when the derivative of $x-x^2$ is $0$, then we have the max that we can have for $x$; $$f'(x-x^2) = 0$$ $$1-2x = 0$$ $$x = 0.5$$
If $x$ is $0.5$, then $y$ obviously is so as well because of $y = 1-x$. 
This then means that $\frac{1}{xy}$ is it's smallest possible value when $x = y = 0.5$, and in that case it's $\frac{1}{0.5^2} = 4$.
Considering that we had the smallest possible value for the $LHS$ be $4$, it must necessarily be greater than or equal to the $RHS$ $4$.

Using the same logic, the second part follows easily. Substitute $y = 1-x$ into the fractions in the second part; we want the $LHS$ to be as small as possible, which means the two fractions added together needs to be as small as possible. This in turn means the denominators of the fractions, $x^2$ and $(1-x)^2$, needs to be as large as possible. Now we maximize $x^2 + (1-x)^2$ in the same way by taking the derivative and setting it to $0$:
$$f'(x + (1-x)^2) = 0$$
$$4x - 2 = 0$$
$$x = 0.5$$
So the smallest value that $(1 + \frac{1}{x^2})(1 + \frac{1}{y^2})$ can have is when $x = 0.5$, and similarly $y = 1-0.5 = 0.5$. Seeing what this gives us we get that $(1 + \frac{1}{0.5^2})(1 + \frac{1}{0.5^2}) = 25$, which of course is larger than $2$, considering that the smallest value the $LHS$ can have is $25$, the inequality must necessarily always hold.
A: Hint
$(x-y)^2\geq 0$
$(x-y)^2 +4xy\geq 4xy $
$(x+y)^2\geq 4xy$
$1\geq 4xy$
