Is the relation $P$, for all real numbers $x$ and $y$ that satisfy $xPy $ iff $x^3 - y \ge y^3 - x$, a reflexive, symmetric and transitive relation? Image of an exam question I am revising link: [1]
For (i) I have stated the relation is reflexive as $\forall x ∈ \Bbb R, xPx$ is reflexive as $x^3 \ge x $
For (ii) I have stated that the relation is not symmetric as $\forall x,y ∈ \Bbb R, xRy$ does not equal $yRx$, as for the values $x=2, y=1, x^3 - y \ge y^3 - x$ does not equal to $y^3 - x \ge x^3 - y$
For (iii) I have stated that the relation is transitive as as $xPy$ and $yPz$ imply $xPz \forall x,y,z ∈ P,$ as $\forall x,y,z$ that are elements of $R$ and that satisfy $P$, $(a,b)$ $a$ must always be greater than $b$, therefore $\forall x,y ∈ P x\ge y, \forall y,z ∈ P y\ge z,$ therefore $x\ge z$
The last one is a tad all over the place, but I just wanted to know if I am on the right track and if my answers are correct, if not I would love to know why so I can improve myself and see where I went wrong as to not make the same mistake in the future. Any and all help would be greatly appreciated.
 A: Your resolution is unclear and incorrect in some places.
For reflexivity, that's not what you have to prove (which is also wrong! as, if $x=0.5$ then $x^3\not\ge x$), you have to show that $x^3-x\ge x^3 -x$ which is obviously true (as the expressions are equal).
Also, saying "$xRy$ does not equal $yRx$" doesn't make much sense, you should use a word like equivalent to describe that two propositions imply the same things.

Here's perhaps an easier resolution.
Note that
$$xPy\iff x^3-y\ge y^3-x\iff x^3+x\ge y^3+y.$$
From that it's clear that the relation is reflexive (explained above).
Now, Let $x=1,y=0$, we have that $2=x^3+x\ge y^3+y=0$, and so $xPy$, but $yPx$ is not true (why?), thus $P$ is not symmetric (in fact, it's asymmetric!).
Now, for transitivity, suppose $xPy$ and $yPz$, this says that $x^3+x\ge y^3+y$ and $y^3+y\ge z^3+z$, can you follow?
A: An easier way to deal with this relation is to define $g(x)=x^3+x$ and say that $xPy\iff g(x)\ge g(y)$. Which immediately shows that $P$ is indeed transitive but not symmetric.
