10.3 Proving Inequalities via Transitivity

As we’ve seen, when proving an inequality, one option is to create a sequence of equivalent inequalities and end up with an inequality you already know is true (using the transitivity of equivalent statements). Another option is to use the transitivity of the inequality itself. If we want to show $a\leqslant b$ , it’s enough to show that $a\leq c$ and $c\leq b$ .

Example 10.16.

Suppose we wish to prove that

\frac{n^{3}+2n^{2}-n}{n^{2}+n}\leq 3n

for all $n\in\mathbb{N}$ . If we construct a sequence of upper bounds for $\frac{n^{3}+2n^{2}-n}{n^{2}+n}$ , with each new bound being an upper bound of the previous bound, ending with $3n$ as our final upper bound, we will be able to use the transitivity of $\leq$ to prove our desired inequality.

Indeed, observe that, for all $n\in\mathbb{N}$ ,

\frac{n^{3}+2n^{2}-n}{n^{2}+n}=\frac{n^{2}+2n-1}{n+1}\leq\frac{n^{2}+2n}{n+1},

since $n^{2}+2n-1\leq n^{2}+2n$ . Also,

\frac{n^{2}+2n}{n+1}\leq\frac{n^{2}+2n}{n}=n+2,

since $n+1\geq n$ (remember that we must decrease our denominator to increase the size of our overall fraction). Finally, since $2n\geq 2$ for all $n\in\mathbb{N}$ , we have

n+2\leq n+2n=3n.

Thus,

\frac{n^{3}+2n^{2}-n}{n^{2}+n}\leq 3n

for all $n\in\mathbb{N}$ as stated.

Example 10.17.

Suppose we wish to prove that

\left|\frac{n-3}{n^{2}+2n-4}\right|\leq\frac{1}{n}

for all integers $n\geq 3$ . Notice that if $n\geqslant 3$ , we get $n-3\geqslant 0$ and $n^{2}+2n\geqslant 3^{2}+2(3)=15\geqslant 4$ . Thus, for such $n$ , $\frac{n-3}{n^{2}+2n-4}\geqslant 0.$ Hence,

\left|\frac{n-3}{n^{2}+2n-4}\right|=\frac{n-3}{n^{2}+2n-4}\leqslant\frac{n}{n^% {2}+2n-4}\leq\frac{n}{n^{2}+2n-2n}=\frac{1}{n},

for all $n\geq 3$ .