$f(xy + x +y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$. 
Let $f : \mathbb R \to \mathbb R$ that satisfies both 2 conditions ,
$f(xy + x + y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$ $\forall x,y \in \mathbb R$.
Determine all such $f$.

My solution.
Let $P(x,y) $ be $f(x)(y - x) + f(y)(x - y) \geq 0$.
Let $Q(x,y)$ be $f(xy + x + y) = f(xy) + f(x) + f(y)$.
First , I show that $f$ is decreasing $\forall x \in \mathbb R$
Proof : Let $x \gt y$ , from $P(x,y)$ we get that $f(x) \leq f(y)$.  (Because $(x - y) \leq 0)$
So, $f$ in decreasing.
$Q(0,0)$ $\to$ $f(0) = 3f(0)$ $\to$ $f(0) = 0$.
$Q(x,-x)$ $\to$ $f(-x^2) = f(-x^2) + f(x) + f(-x)$. $\to$ $f(x) = -f(-x)$.
$P(x,2x)$ $\to$ $xf(x) \geq xf(2x)$. $\to$ $f(x) \geq f(2x)$.
Since $f(x) \geq f(2x)$ , take $x \lt 0$ , we get that $f$ is increasing $\forall x \lt 0$. This implies $f(x) = c$ and $f(x) = -c$ ; $c \in \mathbb R$.
$P(x,y)$ when $x,y \gt 0$ implies $-c = -3c$ , $c = 0$.
So, $f(x) = 0$ $\forall x \in \mathbb R$.   Q.E.D.
Is my proof correct? (This is from 2016 Thailand POSN Camp 2)
 A: You can show that solutions to the functional equation
$$ f ( x y + x + y ) = f ( x y ) + f ( x ) + f ( y ) \tag 0 \label 0 $$
are exactly the additive functions, i.e. those satisfying
$$ f ( x + y ) = f ( x ) + f ( y ) \text . \tag 1 \label 1 $$
It's easy to see that additive functions satisfy \eqref{0}. I try to show the converse, repeating some of your own arguments, for the sake of completeness. First, letting $ x = y = 0 $ in \eqref{0} we get $ f ( 0 ) = 0 $. Then, plugging $ y = - 1 $ in \eqref{0} we find out that
$$ f ( - x ) = - f ( x ) \text . \tag 2 \label 2 $$
Substituting $ - x $ for $ x $ in \eqref{0} and using \eqref{2} we get $ f ( y - x - x y ) = f ( y ) - f ( x ) - f ( x y ) $, which together with  \eqref{0} gives
$$  f (  x y + x + y ) + f (  y - x - x y ) = 2 f ( y ) \text . \tag 3 \label 3 $$
For $ y \ne - 1 $, we can substitute $ \frac { x - y } { y + 1 } $ for $ x $ in \eqref{3} and see that $ f ( x ) + f ( 2 y - x ) = 2 f ( y ) $, hence $ f ( 2 y ) = 2 f ( y ) $ by letting $ x = 0 $, which transforms the previous equation to $ f ( x ) + f ( 2 y - x ) = f ( 2 y ) $. Substituting $ \frac { x + y } 2 $ for $ y $ in the last equation, we get \eqref{1} for $ x + y \ne - 2 $. Substituting $ - x $ for $ x $ and $ - y $ for $ y $ and using \eqref{2}, we'll have \eqref{1} for $ x + y \ne 2 $, which shows that \eqref{1} holds for every $ x $ and $ y $.
Finally, it is well-known that monotone additive functions are exactly those  of the form $ f ( x ) = c x $, for some constant $ c $. The decreasing ones are exactly those with $ c \le 0 $, which form the class of all solutions of \eqref{0} satisfying
$$ f ( x ) ( y - x ) + f ( y ) ( x - y ) \ge 0 \text . $$
A: Putting $y = 1$,
$f(x + x + 1) = f(x) + f(x) + f(1)$
$f(2x + 1) = 2f(x) + f(1)$
Under the assumption that $f$ is continuously differntiable, you can say the following:
$2f'(2x+1) = 2f'(x) \implies f'(2x+1) = f'(x)$
This gives us,
$f'(x) = f'(2(\frac{x-1}{2}) + 1) = f'(\frac{x-1}{2})$
$f'(\frac{x-1}{2}) = f'(2(\frac{x}{4} - \frac{1}{4} - \frac{1}{2}) + 1) = f'((\frac{x}{4} - \frac{1}{4} - \frac{1}{2}))$
In general,
$f'(x) = f'(\frac{x}{2^n} - \frac{1}{2^n} - \frac{1}{2^{n-1}} - ...-\frac{1}{2})$
That is,
$f'(x) = f'(\frac{x}{2^n} - \frac{2^{n-1} - 1}{2^n})$
Because of continuity of $f'$, we have
$\lim_{n \to \infty}f'(x) = \lim_{n \to \infty}f'(\frac{x}{2^n} - \frac{2^{n} - 1}{2^n})$
$f'(x) = f'(0 - 1 + \frac{1}{2^n}) = f'(-1)$
So, $f'(x) = f'(-1) \implies f(x) = xf'(-1) + c$
Recall, $f(2x + 1) = 2f(x) + f(1)$, putting $x = 0$, we have,
$f(1) = 2f(0) + f(1) \implies f(0) = 0$, so $c = 0$, in above.
$f'(x) = f'(-1)x$, by second condition you have already shown that $f'$ is -ve. So, I wonder if we can say $f'(x) = -cx, c > 0$.
Now, I know that the assumption of continuity and differentiability is a big one and it is quite possible that the given conditions could be used to get there or I could be completely wrong altogether. In any case, I thought I would update what I had so far.
