How to demonstrate this inequality? I have this statement:

If $x > 0 , y > 0$, prove that $\frac{1}{x}+\frac{1}{y} > \frac{2}{x+y}$

I will get the hypothesis? (That's what my teacher calls him, I do not know if it's correct)
$\frac{1}{x}+\frac{1}{y} - \frac{2}{x+y}> 0$
$\frac{x^2 + y^2 + xy + xy - 2xy}{(x+y)xy} > 0$
The hypothesis: $\frac{x^2 + y^2}{(x+y)xy} > 0$
Now I am going to prove, i mean, I need to go back to the origin, which is:
$ \frac {1} {x} + \frac {1} {y}> \frac {2} {x + y} $
Here is my problem, and it is hard for me to go back to the origin, step by step, could you help me? I have 5 sheets worn.
PD: I tried this:

 A: Observe
\begin{align}
\frac{1}{x}-\frac{1}{x+y}+\frac{1}{y}-\frac{1}{x+y}>\frac{1}{x}-\frac{1}{x}+\frac{1}{y}-\frac{1}{y}=0
\end{align}
A: Use a better hypothesis, and get a better inequality.
\begin{align}
\frac{(x-y)^2}{(x+y)xy} &\ge 0 \\
\frac{x^2 - 2xy + y^2}{(x+y)xy} &\ge 0 \\
\frac{x^2 + 2xy + y^2}{(x+y)xy} &\ge \frac{4xy}{(x+y)xy} \\
\frac{(x+y)^2}{(x+y)xy} &\ge \frac{4}{x+y} \\
\frac{x+y}{xy} &\ge \frac{4}{x+y} \\
\frac{1}{x} + \frac{1}{y} &\ge \frac{4}{x+y} > \frac{2}{x+y} \\
\end{align}
A: Hint:   by the AM-HM inequality $\;\;\displaystyle \frac{x+y}{2} \,\ge\, \frac{2}{\cfrac{1}{x}+\cfrac{1}{y}} \,\gt\, \frac{1}{\cfrac{1}{x}+\cfrac{1}{y}}\,$.
A: You made it too complex: as all numbers are positive, you just have to prove that
$$\frac{x+y}{xy}>\frac2{x+y}\iff (x+y)^2>2xy,$$
which is obvious since $(x+y)^2=x^2+y^2+2xy$.
A: The other answers give alternative solutions to your problem, but your approach also works. I will elaborate on what Brian Tung was saying in his comment to your question.
These are the algebraic steps you took:
$$\dfrac{1}{x} + \dfrac{1}{y} > \dfrac{2}{x+y}$$
$$\dfrac{1}{x} + \dfrac{1}{y} - \dfrac{2}{x+y} > 0$$
$$\dfrac{y(x+y)}{xy(x+y)} + \dfrac{x(x+y)}{yx(x+y)} - \dfrac{2xy}{(x+y)xy} > 0$$
$$\dfrac{y(x+y) + x(x+y) - 2xy}{(x+y)xy} > 0$$
$$\dfrac{xy + y^2 + x^2 + xy - 2xy}{(x+y)xy} > 0$$
$$\dfrac{x^2+y^2}{(x+y)xy} > 0$$
Basically, you will want to write these algebraic steps backwards, as follows:
Given $x>0$ and $y>0$, we have that
$$\dfrac{x^2+y^2}{(x+y)xy} > 0 \tag{1}$$
$$\dfrac{xy + y^2 + x^2 + xy - 2xy}{(x+y)xy} > 0 \tag{2}$$
$$\dfrac{y(x+y) + x(x+y) - 2xy}{(x+y)xy} > 0 \tag{3}$$
$$\dfrac{y(x+y)}{xy(x+y)} + \dfrac{x(x+y)}{yx(x+y)} - \dfrac{2xy}{(x+y)xy} > 0 \tag{4}$$
$$\dfrac{1}{x} + \dfrac{1}{y} - \dfrac{2}{x+y} > 0 \tag{5}$$
$$\dfrac{1}{x} + \dfrac{1}{y} > \dfrac{2}{x+y} \tag{6}$$
However, you can't just do this reversal blindly. You need to check to make sure that each step in the resulting sequence of steps is valid. (Essentially, that all of your original steps are invertible.)
To get from $(1)$ to $(2)$, you add $xy+xy-2xy$ to the numerator, which is fine because $xy+xy-2xy = 0$, and adding 0 doesn't change an expression. Getting from $(2)$ to $(3)$ is simply reordering terms and then factoring, which is another valid step. $(3)$ to $(4)$ is just breaking apart a fraction into a sum of fractions; still valid. $(4)$ to $(5)$ is also valid, since $x$, $y$, and $x+y$ are all non-zero, so we can divide both the numerator and denominator by any of them. Finally, $(5)$ to $(6)$ is valid because you can add any term to both sides of an inequality.
Checking these steps may seem a bit pedantic, but it's important for some problems. For example: "If $x^2 = 1$, prove $x = 1$". This isn't actually true, because if $x^2=1$, then $x$ can instead equal $-1$. But if you start from $x=1$ and square both sides , you get $x^2=1$. The problem is that the inverse operation, taking the square root of both sides, requires putting a $\pm$ on one of the sides.
Strictly speaking, the algebraic steps you took to get from $\dfrac{1}{x} + \dfrac{1}{y} > \dfrac{2}{x+y}$ to $\dfrac{x^2+y^2}{(x+y)xy}$ don't directly contribute to your proof (shameless plug: I wrote an answer to another question that goes into the logical details about why). Rather, those steps serve as a heuristic to find your real solution, which usually works since most algebraic steps are invertible. The reason for using such a heuristic process is that it's easy to notice where we need to simply expressions, but it's hard to see (in a psychological sense) where we'd need to unsimplify expressions, which is what our final proof steps ended up being for the most part.
A: An option:
Let: $x,y >0.$
Note: 
1)$\dfrac{1}{x} \gt \dfrac{1}{x+y}.$
2) $\dfrac{1}{y} \gt \dfrac{1}{x+y}.$
Hence:
$\dfrac{1}{x}+\dfrac{1}{y} \gt \dfrac{2}{x+y}.$
Note: On the RHS of 1) ,2) the denominator has been increased by a positive number, the fraction decreased.
A: I'm going to use your hypothesis and continue from there.
Please note that this is by no means a proper mathematical proof but merely a simple consequence from the given. When I saw the problem this is what I first thought of but seen that it wasn't really mentioned.
So we have that $$\frac{x^2+y^2}{(x+y)xy} \gt 0$$
We were also originally given that $x\gt 0$ and $y \gt 0$. This means that $x$ and $y$ must be positive numbers! 
If we look at how positive numbers work, we see that the square of any positive number is positive. And we can also see that the sum of two positive numbers is always positive. Thus our numerator is some positive value. The denominator is also guaranteed to be positive because multiplying any three positive numbers gives a positive number.
Dividing two positive numbers also always gives a positive number, even if it is one that is really really small. Since it's positive, it's still bigger than $0$.
