How to find the sup, inf, max and min for the set $H=\{ x \in \Bbb R : x+\frac3x-1 \le 0\}$ Let $H=\{ x \in \Bbb R : x+\frac3x-1 \le 0\}$.  Find $\sup(H)$, $\max(H)$, $\inf(H)$, and $\min(H)$. I understand the terminology but I am new to this please just offer any first steps or any help on how to get started.
As the expression on the left is continuous, the values of $x$ for which it is true will form one or more intervals.  The transitions will either be where $x+\frac 3x-1=0$ or where it is undefined, so solve the equation, add $0$ to the set of roots because the expression is undefined there, and assess the truth of the inequality on each resulting interval.  Can you then find the $\inf, \sup, \max, \min$ once you have the interval(s)?
Notice that we have 
$$x+\frac3x \ge 2\sqrt3$$
for $x&gt;0$. There are various ways to show this, you can have a look at several posts with proofs of 
similar inequality $x + \frac{1}{x} \geq 2 $.

Let us denote $f(x)=x+\frac3x-1$. For $x&gt;0$ you have
$$f(x) = x+\frac3x-1 \ge 2\sqrt3-1 &gt;0.$$
And for $x&lt;0$ you have
$$f(x) = -\left(|x|+\frac3{|x|}\right) -1 \le -2\sqrt3 - 1 &lt;0.$$

So you can see that $f(x)&gt;0$ for positive values and $f(x)&lt;0$ for negative values. (You can also try to sketch the graph to see this.) 

Which basically means that
$$H=\{x\in\mathbb R; f(x)\le 0 \} = \{x\in\mathbb R; x&lt;0\}.$$
Now that you have much simpler description of $H$, it should be easier to find supermum, infimum, maximum, minimum of $H$ (if they exist).