Hi I am a student working through Spivak Calculus chapter 7 theorem 9 pg.123 and I can't understand his use of the triangle inequality in his theorem.
The theorem states: If $n$ is odd, then any equation $$ \ x^n+a_{n-1}x^{n-1} +\cdots+a^0 $$ has a root.
proof: we would like to prove that $f$ is sometimes positive and sometimes negative. The intuitive idea is that for large $|x|$, the function is very much like $g(x) = x^n$ and, since $n$ is odd, this function is positive for large positive $x$ and negative for large negative $x$. A little algebra is all we need to make this intuitive idea work.
$$ f(x) = x^n+a_{n-1}x^{n-1} +\cdots+a^0 = x^n \left(1+\frac{a_{n-1}}{x}+\cdots+\frac{a_0}{x^n}\right) $$
Note that $$ \left|\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right|\le \frac{|a_{n-1}|}{|x|}+\frac{|a_{n-2}|}{|x^2|}+\cdots+\frac{|a_{0}|}{|x^n|} $$
Consequently if we choose $x$ satisfying $$ |x|>1,2n|a_{n-1}|,\ldots,2n|a_0| $$
Then
$$ \left|\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right|\le \frac{1}{ 2n}+\frac{1}{2n}+\cdots+\frac{1}{2n}=\frac{1}{2n}$$ what i do not understand is how can we say that the summation of these terms is still $\le\frac{1}{2n}$ surely it is strictly less than $\frac{1}{2n}$ and never equal to as I am taking x to be smaller?.