Adding Inequalities

Combining Inequalities by Addition

The previous tutorial established how inequalities behave under translation, negation, and multiplication by a number of known sign. A further useful operation is to add two inequalities. This lets us combine separate bounds, such as a lower bound for one number and a lower bound for another. The comparisons must point in compatible directions: adding a smaller quantity to a smaller quantity gives a comparison between the sums.

Throughout this tutorial, $F$ denotes an ordered field. The arguments therefore apply in $\mathbb{R}$, as well as in any other ordered field. Recall that $a\leq b$ means either $a<b$ or $a=b$. The result of adding inequalities depends on whether either input comparison is strict: two non-strict comparisons give a non-strict result, and a strict comparison among them makes the resulting comparison strict.

Theorem (Addition of inequalities). Let $a,b,c,d\in F$. If $a\leq b$ and $c\leq d$, then $$ a+c\leq b+d. $$ If at least one of the original comparisons is strict, then $a+c<b+d$. In particular, if $a<b$ and $c<d$, then $a+c<b+d$.

Proof. Since $a\leq b$, either $a=b$ or $a<b$. Since $c\leq d$, either $c=d$ or $c<d$. If both are equalities, then $a+c=b+d$. If $a=b$ and $c<d$, translation of $c<d$ by $a$ gives $a+c<a+d=b+d$. If $a<b$ and $c=d$, translation of $a<b$ by $c$ gives $a+c<b+c=b+d$. Finally, if both comparisons are strict, translating $a<b$ by $c$ gives $a+c<b+c$; translating $c<d$ by $b$ gives $b+c<b+d$. Transitivity of the strict order then gives $a+c<b+d$. These cases prove both the non-strict conclusion and the claim about strictness. $\square$

The theorem concerns two whole comparisons. Each left-hand side is added to the other left-hand side, and each right-hand side is added to the other right-hand side. The rule does not say that any four expressions can be arranged in a useful comparison merely because they appear in inequalities. Keeping track of which quantities are being added is essential.

Worked Example: Adding Lower Bounds

Suppose $x\geq 2$ and $y\geq -5$. In the standard left-to-right order, these statements are $2\leq x$ and $-5\leq y$. Adding the corresponding sides gives

2+(-5)\leq x+y, \qquad\text{so}\qquad -3\leq x+y.

The conclusion allows equality. For instance, if $x=2$ and $y=-5$, both original bounds are equalities and $x+y=-3$. If either original bound is strict, the addition theorem instead gives $-3<x+y$. Thus one should not claim a strict final bound unless at least one input comparison is strict.

When Is the Sum Comparison Strict?

The equality cases in the proof give a precise way to remember the rule. When both comparisons are non-strict, the sum can be equal only when both input comparisons are equalities. If even one input has a positive gap between its sides, adding the comparisons preserves a positive gap in the total.

Corollary (Equality case for adding inequalities). Suppose $a\leq b$ and $c\leq d$. Then $a+c=b+d$ if and only if $a=b$ and $c=d$.

Proof. If $a=b$ and $c=d$, then addition gives $a+c=b+d$. Conversely, suppose $a+c=b+d$. The addition theorem gives $a+c\leq b+d$, and the assumed equality rules out a strict comparison. In the case analysis in the proof of the theorem, equality occurs only in the case $a=b$ and $c=d$. Therefore both original comparisons are equalities. $\square$

Equivalently, among two non-strict comparisons, a strict input makes the sum comparison strict. This is useful when combining bounds: if one estimate is exact at its endpoint but another is strict, the combined estimate is still strict. The equality case also cautions against inferring equality of each part from a sum without first knowing that both parts satisfy the required inequalities.

Worked Example: A Strict Bound for a Sum

Suppose $p>4$ and $q\geq -2$. Rewrite these as $4<p$ and $-2\leq q$. Add the inequalities in that order:

4+(-2)<p+q, \qquad\text{so}\qquad 2<p+q.

The result is strict because $4<p$ is strict, even though $-2\leq q$ allows equality. To check the endpoint behavior, if $p=5$ and $q=-2$, then $p+q=3>2$. If $p=4$ and $q=-2$, the original strict hypothesis $p>4$ fails; that equality case cannot be used to weaken the conclusion.

Combining Bounds on Several Quantities

A common use of addition is to turn separate bounds into a bound on a sum. For lower bounds, write each comparison with the lower bound on the left. For upper bounds, write each comparison with the upper bound on the right. The addition theorem then combines the lower bounds with each other and the upper bounds with each other.

For example, if $L\leq x\leq U$ and $M\leq y\leq V$, then the lower comparisons $L\leq x$ and $M\leq y$ give $L+M\leq x+y$. The upper comparisons $x\leq U$ and $y\leq V$ give $x+y\leq U+V$. Together they bound the sum from below and above. Either resulting comparison becomes strict if one of its two component comparisons is strict.

Worked Example: Bounding a Sum on Both Sides

Suppose $-3\leq r\leq 6$ and $2\leq s<5$. For the lower bound, add $-3\leq r$ and $2\leq s$:

-3+2\leq r+s, \qquad\text{so}\qquad -1\leq r+s.

For the upper bound, add $r\leq 6$ and $s<5$. The second comparison is strict, so the sum comparison is strict:

r+s<6+5, \qquad\text{so}\qquad r+s<11.

Consequently, $-1\leq r+s<11$. The lower endpoint is non-strict because both lower comparisons allow equality. The upper endpoint is strict because $s<5$. Both details follow directly from the equality and strictness cases of the addition theorem.

Adding Inequalities to Compare Differences

Subtraction can also be handled using addition. The order-reversal-under-negation theorem from the previous tutorial says that $c\leq d$ implies $-d\leq -c$. Adding this reversed comparison to $a\leq b$ gives a comparison of differences. The direction of the second comparison must be reversed before it is added.

Theorem (Comparison of differences by addition). If $a\leq b$ and $c\leq d$, then $$ a-d\leq b-c. $$ If at least one of the two original comparisons is strict, then $a-d<b-c$.

Proof. From $c\leq d$, order reversal under negation gives $-d\leq -c$. We can therefore add $a\leq b$ and $-d\leq -c$, applying the addition theorem:

a+(-d)\leq b+(-c).

By the definition of subtraction, this is $a-d\leq b-c$. If $a<b$, the first input comparison is strict. If $c<d$, negation gives $-d<-c$, so the second comparison is strict. In either case the strictness clause of the addition theorem gives $a-d<b-c$. $\square$

Worked Example: Comparing a Difference

Suppose $m\leq 8$ and $3\leq n$. To obtain a bound on $m-n$, reverse the second comparison by negating its sides: $n\geq 3$ gives $-n\leq -3$. Now add $m\leq 8$ and $-n\leq -3$:

m+(-n)\leq 8+(-3), \qquad\text{so}\qquad m-n\leq 5.

The direction matters: the lower bound $n\geq 3$ becomes an upper bound $-n\leq -3$ after negation. If instead $m<8$, the final comparison is strict, so $m-n<5$. If $n>3$, negating gives $-n<-3$, and the final comparison is strict for that reason as well.

Why the Direction of Each Inequality Matters

The addition rule requires the inequalities to be oriented consistently: smaller quantities on the left and larger quantities on the right. If one comparison points the other way, reverse it first when appropriate, using order reversal under negation or simply rewriting the comparison in increasing order. Do not add the left and right sides without checking the direction of both input comparisons.

For instance, $1<x$ and $x<4$ point in the same direction, so the addition theorem gives $1+x<x+4$, which simplifies to $1<4$. This is true, but it does not produce a bound on $x$: the $x$ appears once on each side and cancels. To produce a stronger statement such as $2<2x$, add $1<x$ to itself. That gives $1+1<x+x$, hence $2<2x$. The expressions being added determine what conclusion follows.

A useful discipline is to write out the two comparisons and their sums before simplifying. Check that the left sides are added together, the right sides are added together, and that the strictness of the conclusion is justified. This avoids two common errors: adding inequalities that point in opposite directions without first changing their orientation, and replacing a valid non-strict conclusion with a strict one when both input comparisons could be equalities.

Key takeaway. If $a\leq b$ and $c\leq d$, then $a+c\leq b+d$. The result is strict whenever either input comparison is strict. To compare differences, first reverse one inequality by negation, then apply the same addition rule.

Check Your Understanding

Use the addition theorem and the order-reversal-under-negation theorem to answer the following questions.

If $u\leq 5$ and $v\leq -2$, what upper bound follows for $u+v$?
If $a<b$ and $c\leq d$, is the comparison between $a+c$ and $b+d$ strict? Explain.
Under the assumptions $a\leq b$ and $c\leq d$, when can $a+c=b+d$ hold?
If $x\leq 9$ and $4\leq y$, what comparison follows for $x-y$?
What should you check before adding two inequalities, and why does the strictness of an input matter?

What You'll Learn