<s>[INST] 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.
 [/INST] 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 $x^4+1\geq x^3+x$.
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