Help with discrete math proof? I am having trouble proving the following:
If $x\in R$ and $x > 0$, then $x^4+1 \geq x^3+x$. 
Work: I tried to rearrange the equation as $x^4-x^3-x+1 \geq 1$, but that does not really help. I also tried proof by cases where case 1 would be that x is irrational and case 2 would be that x is rational. However, that has not got me far either. I am not really sure how to approach this problem.
 A: Hint: $x^4 - x^3 - x + 1 = x^3(x - 1) - (x - 1) = (x^3 -1)(x-1) = (x-1)^2 (x^2 + x + 1)$.
A: Hint: try the rearrangement inequality.
details: if $x>1$:
$$
\frac 1x < 1\\
x<x^4\\
x + x^3 = 1\cdot x + \frac 1x \cdot x^4 \le \frac 1x \cdot x + 1\cdot x^4 = 1+x^4
$$
Otherwise:
$$
\frac 1x \ge 1\\
x\ge x^4\\
x + x^3 = 1\cdot x + \frac 1x \cdot x^4 \le \frac 1x \cdot x + 1\cdot x^4 = 1+x^4
$$
A: As $x\in \mathbb{R}$ and $x>0$, then we will analyze the sign of the expression
$$x^4-x^3-x+1. $$
Indeed,
$$x^4-x^3-x+1=x^3(x-1)-1(x-1)=(x-1)(x^3-1) $$
Now, as $x^3-1=(x-1)(x^2+x+1)$, then
$$x^4-x^3-x+1=(x-1)(x-1)(x^2+x+1) =(x-1)^2(x^2+x+1).$$
Note that $(x-1)^2(x^2+x+1)\geq 0$ because $x>0$ (it can be zero if $x=1>0$). 
Therefore, it follows that $x^4-x^3-x+1\geq 0$, then $x^4+1\geq x^3+x$.
A: Rearrange as you say and note that
$$x^4 - x^3 - x +1 = (x-1)(x^3 -1) =  (x-1)(x-1)(x^2+x+1) = (x-1)^2(x^2+x+1).$$
This first factor is always non-negative. So is the second, as the descriminant of this quadratic factor is negative, and the coefficient of the leading term in the quadratic factor is positive (so that the graph of the quadratic factor is a "U"-shape that doesn't intersect the $x$-axis).
The product is therefore non-negative $\forall x \in \mathbb{R}$.
