Spivak Calculus Prologue I'm completely blown away by the difficulty of Spivak. I've managed to work through the first 3 problems, but I feel I'm missing something important to solve these basic inequalities in his 4th problem:
$$x^2 + x + 1 > 2$$
&
$$x^2 + x + 1 > 0$$
any suggestions?
 A: Complete the square:
$$
x^2 + x + 1 = (x^2 + x + \tfrac 1 4) + \frac 3 4 = \left(x + \frac 1 2 \right)^2 + \frac 3 4.
$$
That gets you the second one.
For the first one, put everything on one side of the inequality and $0$ on the other side, and procede similarly.
Later addendum in response to vitno's question in the comments below:
In general, the process of completing the square looks like this:
$$
\begin{align}
ax^2 + bx + c & = a\left(x^2 + \frac b a x\right) + c \\[12pt]
& = a\left(x^2 + \frac b a x + \frac{b^2}{4a^2}\right) + c - a\left(\frac{b^2}{4a^2}\right) \tag{$\begin{array}{c} \text{completing} \\  \text{the square}\end{array}$} \\[12pt]
& = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a}.
\end{align}
$$
Say you have a particular case:
$$
3x^2 + 20 x + 7.
$$
Proceed as follows:
$$
3\left(x^2 + \frac{20}{3} x \right) + 7.
$$
Take half the coefficient of the first-degree term and square it, getting $(10/3)^2$.  Add this in the appropriate place, and substract it out later:
$$
3\underbrace{\left(x^2 + \frac{20}{3} x + \left(\frac{10}{3}\right)^2 \right)}_{\text{a perfect square}} + 7 - 3\left(\frac{10}{3}\right)^2
$$
$$
= 3\left(x + \frac{10}{3}\right)^2 - \frac{79}{3}.
$$
Knowing how and when to complete the square is useful.
Remember this: The purpose of completing the square is always to reduce a quadratic polynomial with a first-degree term to a quadratic polynomial with no first-degree term.
A: $x^2+x+1>2 \iff x^2+x+\frac{1}{4}>\frac{5}{4}$
Thus, $(x+\frac{1}{2})^2\gt\frac{5}{4}$.  We can reduce this to $|x+\frac{1}{2}|>\frac{\sqrt{5}}{2}$.  Thus, our final answer will be
$x<-\frac{\sqrt{5}+1}{2}$ or $x>\frac{\sqrt{5}-1}{2}$.
For the second answer, notice that we can apply the same strategy to the previous problem and get $(x+\frac{1}{2})^2>-\frac{3}{4}$.  Notice, that if $x$ is a real number, then squaring $(x+\frac{1}{2})$ will never create a negative number, so this inequality holds for all real numbers.
A: Hint: start by considering the graph of $y = x^2+x+1$.  What is its shape?  Where is $y=0$ or $2$?
A: We have $x^2+x+1=(x+1/2)^2+3/4$. Now any inequalities you want to prove involve only a square.
So for example for $x^2+x+1 \gt 2$, rewrite as $(x+1/2)^2\gt 5/4$, giving $|x+1/2|\gt (1/2)\sqrt{5}$, which can be rewritten in various ways.
For $x^2+x+1 \gt 0$, note that in fact $x^2+x+1 \ge 3/4$, since any square is $\ge 0$.  
