$f(xy + x +y) = f(xy) + f(x) + f(y)$ 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)

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.
