Triangle Inequality

From Two Terms to Finite Sums

The Triangle Inequality, established in the Absolute Value tutorial, says that \(|x+y|\leq |x|+|y|\) for real numbers \(x\) and \(y\). Its practical strength is that it bounds a combined quantity without requiring us to know its sign. This tutorial extends that idea to any finite list of real numbers. The extension is useful when an expression has several terms, or when several separate errors contribute to one total.

The key is to apply the two-term inequality repeatedly. For example, first group the first two terms, then combine their sum with the third term. At each stage, the absolute value of the partial sum is bounded by the sum of the absolute values already included. This gives an upper bound even if some terms cancel one another.

Proof. We use induction on \(n\). If \(n=1\), then the two sides are both \(|x_1|\), so the inequality holds. Now suppose it holds for some positive integer \(n\). For real numbers \(x_1,\ldots,x_n,x_{n+1}\), the Triangle Inequality for two real numbers gives

By the induction hypothesis, \(\left|\sum_{k=1}^{n}x_k\right|\leq\sum_{k=1}^{n}|x_k|\). Adding \(|x_{n+1}|\) to both sides preserves the inequality. Therefore,

This proves the statement for \(n+1\), and induction proves it for every positive integer \(n\). \(\square\)

The theorem says that the magnitude of a sum cannot exceed the sum of the magnitudes of its terms. It does not say that the two quantities are always equal. When terms have different signs, their contributions can cancel in the sum, while the right-hand side counts every magnitude as nonnegative.

Using the Inequality with Grouped Terms

A finite sum can be grouped before applying the Triangle Inequality. This is useful when some parts of an expression are easier to handle together. For example, the two-term theorem gives \(|a+b|\leq|a|+|b|\) for one group, and it can then be applied to the sum of that group and another term. The finite-sum theorem guarantees that this repeated process is valid no matter how many terms are present.

The bound also applies when the terms are fractions, powers, or values of functions, provided each term is a real number. It is often useful to choose whether to calculate the sum exactly or to bound it directly. If the individual terms are easy to estimate but the sum is awkward, the inequality may give the information needed with less calculation.

When Is the Bound Exact?

The finite-sum inequality becomes an equality precisely when cancellation does not reduce the magnitude of the sum. For real numbers, that happens when all the terms are nonnegative or all are nonpositive. Zeros are allowed in either case: they contribute nothing to the sum of the magnitudes and do not oppose any other term.

Proof. First suppose that all the terms are nonnegative. Then their sum is nonnegative, so its absolute value equals the sum itself. Also \(|x_k|=x_k\) for each \(k\). Consequently,

If instead all the terms are nonpositive, then their sum is nonpositive. Its absolute value is the negative of that sum, and \(|x_k|=-x_k\) for each \(k\). Hence

This proves equality whenever the terms have the same weak sign, including the case in which every term is zero.

Conversely, suppose that

Let \(P\) be the sum of the positive terms among \(x_1,\ldots,x_n\), taking \(P=0\) if there are no positive terms. Let \(N\) be the sum of the magnitudes of the negative terms, taking \(N=0\) if there are no negative terms. Both \(P\) and \(N\) are nonnegative. The sum of all the terms is \(P-N\), and the sum of their absolute values is \(P+N\). Our assumed equality is therefore

If \(P\geq N\), then \(|P-N|=P-N\), so the equation implies \(P-N=P+N\), and hence \(2N=0\). Since \(N\geq0\), this gives \(N=0\): there are no negative terms. If \(P<N\), then \(|P-N|=N-P\), so the equation implies \(N-P=P+N\), and hence \(2P=0\). Thus \(P=0\), and there are no positive terms. In either case, all the terms have the same weak sign. This proves the converse and completes the proof. \(\square\)

Bounding the Total of Several Errors

One important use of the finite-sum Triangle Inequality is to control a total error. Suppose real numbers \(e_1,\ldots,e_n\) represent separate signed errors. A positive error and a negative error can cancel in the total, but cancellation may not be reliable or even known in advance. The inequality gives a bound that remains valid regardless of the signs:

For example, if each of four errors has magnitude at most \(0.02\), then the magnitude of their total is at most \(0.08\). In detail, if \(|e_k|\leq0.02\) for each \(k=1,2,3,4\), adding those inequalities gives

Combining this with the finite-sum Triangle Inequality yields \(|e_1+e_2+e_3+e_4|\leq0.08\). This does not claim that the total error actually reaches \(0.08\). It is a guaranteed upper bound, and the equality theorem explains that equality can occur only if the errors have the same weak sign.

A common pitfall is to treat the inequality as an equality without checking signs. For instance, \(|4+(-1)|=3\), not \(4+1=5\). The right-hand side is a safe bound, not necessarily the exact magnitude. When exactness matters, calculate the signed sum or use the equality criterion. When only a guaranteed bound is needed, the finite-sum inequality avoids having to track every cancellation.