Proof by Introducing an Auxiliary Quantity

Why Introduce Another Quantity?

In Proof by Introducing a Definition, a defined term was expanded into the condition that gives it meaning. A related proof technique is to introduce an auxiliary quantity: an extra number or expression that is not the main object of the claim but helps connect what is known to what must be shown. The auxiliary quantity does not change the hypotheses. It gives us another way to organize their information.

For instance, when two real numbers $x$ and $y$ appear together, their difference $x-y$ may be useful even if the claim is about their sum or their squares. The difference is a real number because $x$ and $y$ are real. Its square is nonnegative, and expanding that square can produce terms already present in the claim. The key is to choose an expression whose properties are known and whose algebra is relevant to the goal.

This method is not limited to differences. One might introduce a sum, a product, a translated variable such as $u=x+3$, or a quantity built from several given objects. The choice depends on the structure of the statement. A useful auxiliary quantity is simple enough to control and closely connected to the expressions in the conclusion.

An auxiliary quantity is a tool, not an extra assumption. Define it from the objects already in the problem, state its domain, and use facts that follow from its definition. At the end, translate the result back into the variables of the original claim.

A Practical Strategy

Choosing an auxiliary quantity is a form of planning. Work backward from the conclusion to see what expression would make it easy to establish, then work forward from the hypotheses to see whether that expression can be controlled. This combines the reasoning used in Working Backward From a Conclusion and Working Forward From Hypotheses.

Inspect the target.
Look for repeated expressions, a difference between two sides, or terms that might become a square after a substitution.

Choose and define a quantity.
For example, set $d=x-y$ or $u=x+3$. Make clear that the quantity is determined by the given variables.

Record what is known about it.
Check its domain and any properties that follow directly, such as $d^2\geq0$ for a real number $d$.

Connect it to the original expressions.
Use an identity or substitution to rewrite the relevant terms without losing track of their values.

Finish in the original variables.
Use the established relation to prove the stated conclusion, and check whether equality or boundary cases are included.

The introduction itself does not prove the theorem. If we set $d=x-y$, then we know exactly what $d$ is, but we still need to show how that definition helps. A proof must establish the needed relation, such as $x^2+y^2$ being expressible using $d^2$. Likewise, a substitution such as $u=x+3$ is valid because it assigns a real number to a real input, but the proof must still carry out the algebra connecting $u$ to the claim.

A Difference Reveals a Nonnegative Square

For any real number $d$, the square $d^2$ is nonnegative. When $d=x-y$, expanding $d^2$ gives $x^2-2xy+y^2$. This simple observation produces an inequality relating three expressions that occur frequently together.

Worked Example: Comparing a Sum of Squares with a Product

Theorem. For all real numbers $x$ and $y$, $$ x^2+y^2\geq 2xy. $$

Proof. Let $x,y\in\mathbb R$, and introduce the auxiliary quantity $d=x-y$. Then $d\in\mathbb R$, so $d^2\geq0$. By the definition of $d$, $$ 0\leq d^2=(x-y)^2=x^2-2xy+y^2. $$ Adding $2xy$ to both sides gives $$ 2xy\leq x^2+y^2. $$ This is the desired inequality.

The auxiliary quantity matters because it packages the comparison into a square whose sign is already known. No restriction on the signs of $x$ or $y$ was needed. Equality holds exactly when $d^2=0$, which is equivalent to $d=0$, and then $x-y=0$, so $x=y$. Conversely, if $x=y$, the two sides are equal. Thus equality occurs precisely when the two real numbers are equal.

This proof also illustrates why the domain should be stated. The fact that $d^2\geq0$ applies because $d$ is real. Here that follows from $x,y\in\mathbb R$ and $d=x-y$. If a problem uses a different domain or operation, the property of the auxiliary quantity must be checked rather than assumed.

Use the Quantity That Matches the Goal

Sometimes the original expression becomes simpler after naming a repeated part. This can be useful even when the auxiliary quantity is not itself nonnegative. The purpose may simply be to make a relation visible and reduce clutter. After the calculation, substitute its definition back so that the conclusion is stated in the original variables.

Worked Example: Simplifying an Expression by a Shift

Claim. For every real number $x$, $$ (x+3)^2-(x+1)(x+5)=4. $$

Proof. Let $x\in\mathbb R$, and introduce $u=x+3$. Then $x+1=u-2$ and $x+5=u+2$. Substituting these expressions into the left-hand side gives $$ (x+3)^2-(x+1)(x+5) =u^2-(u-2)(u+2). $$ Using the difference-of-squares identity, $$ (u-2)(u+2)=u^2-4. $$ Therefore $$ u^2-(u^2-4)=4, $$ which proves the claim.

The new variable is not an additional unknown that needs to be solved for. It is a name for the expression $x+3$, and the relations $x+1=u-2$ and $x+5=u+2$ follow from that definition. Choosing the center expression $x+3$ makes the two other factors symmetric around it, so their product takes a compact form.

This example uses an auxiliary quantity to expose structure in an identity; it does not rely on an inequality or on a quantity being positive. Substitutions are especially effective when several expressions differ by the same fixed amount. The proof remains valid because every occurrence is translated consistently.

A Second Inequality from the Same Idea

The preceding theorem can be used to compare a sum of squares and a product. A related argument becomes especially useful when the variables are positive, because division by their product then preserves inequalities. The positivity hypothesis is essential for that division and must not be omitted.

Worked Example: A Ratio Inequality for Positive Numbers

Theorem. If $x,y\in\mathbb R$ with $x>0$ and $y>0$, then $$ \frac{x}{y}+\frac{y}{x}\geq2. $$

Proof. Let $x,y\in\mathbb R$ satisfy $x>0$ and $y>0$. Introduce the auxiliary quantity $d=x-y$. Since $d$ is real, $d^2\geq0$. Also, $xy>0$, so division by $xy$ preserves the inequality: $$ \frac{(x-y)^2}{xy}\geq0. $$ Expanding the numerator and separating the terms gives $$ \frac{x^2-2xy+y^2}{xy} =\frac{x}{y}-2+\frac{y}{x}\geq0. $$ Adding $2$ to both sides yields $$ \frac{x}{y}+\frac{y}{x}\geq2. $$ This proves the theorem.

The condition $xy>0$ justifies the division and also ensures that both denominators in the conclusion are nonzero. Equality holds precisely when $(x-y)^2=0$, which is equivalent to $x=y$. For positive $x$ and $y$, that condition is possible; for example, $x=y=1$ gives equality. The proof does not infer equality merely from the expression being nonnegative: it checks when the square vanishes.

The same auxiliary quantity can serve different goals. In the first inequality, the square $(x-y)^2$ directly gave a comparison between $x^2+y^2$ and $2xy$. Here, dividing that square by a positive product produces the two reciprocal ratios. The algebra is similar, but the domain assumptions and the target statement are different.

Fixed Sums and the Difference from the Mean

An auxiliary quantity can also keep track of how far two numbers are from one another when their sum is known. If the sum is fixed, their difference measures the imbalance between them. An identity involving both the sum and the difference then gives a lower bound for the sum of their squares.

Worked Example: The Smallest Sum of Squares with a Fixed Sum

Proposition. If $x,y\in\mathbb R$ and $x+y=10$, then $$ x^2+y^2\geq50, $$ with equality if and only if $x=y=5$.

Proof. Let $x,y\in\mathbb R$ satisfy $x+y=10$, and define $d=x-y$. Expanding the squares of the sum and difference gives $$ (x+y)^2+(x-y)^2 =(x^2+2xy+y^2)+(x^2-2xy+y^2) =2x^2+2y^2. $$ Thus $$ x^2+y^2=\frac{(x+y)^2+d^2}{2}. $$ Using $x+y=10$ and $d^2\geq0$, we obtain $$ x^2+y^2=\frac{100+d^2}{2}\geq\frac{100}{2}=50. $$ This proves the inequality.

Equality holds exactly when $d^2=0$, or $d=0$. The equations $x+y=10$ and $x-y=0$ then give $2x=10$, so $x=5$ and $y=5$. Conversely, if $x=y=5$, then $x^2+y^2=25+25=50$. This verifies both directions of the equality condition.

The auxiliary difference $d$ records the part that is not determined by the fixed sum. Its square can only add a nonnegative amount to the baseline value $50$. This explains not only why the bound holds, but also why equality occurs when the two numbers are balanced. The same reasoning applies to any fixed real sum $s$: if $x+y=s$, then the identity above gives $x^2+y^2=(s^2+(x-y)^2)/2\geq s^2/2$.

Choosing Carefully and Avoiding Common Errors

An auxiliary quantity is useful only when its definition and properties match the proof. The following distinctions help keep the reasoning valid.

Choice or step	What it provides	What still needs justification
Set $d=x-y$	A real quantity whose square is nonnegative.	Expand $d^2$ correctly and connect it to the claim.
Set $u=x+3$	A shorter name for a repeated expression.	Rewrite every related expression consistently.
Divide by $xy$	A normalized expression involving ratios.	Verify $xy$ is nonzero and has the required sign.
Use a square to prove equality cases	A way to identify when a nonnegative term is zero.	Check both that equality forces the square to vanish and that the resulting condition gives equality.

A frequent mistake is to introduce a symbol and then treat it as if it were independent of the original variables. If $d=x-y$, it is not valid to choose $d$ freely while keeping $x$ and $y$ fixed; its value is already determined. Another mistake is to use a property without checking its hypothesis. For example, dividing an inequality by $xy$ preserves its direction only after the sign of $xy$ is known. In the ratio theorem, $x>0$ and $y>0$ imply $xy>0$, so the division is justified.

The auxiliary quantity should illuminate the proof, not hide a gap. State how it is defined, verify the property used, and show the algebra that links it to the original goal. If the definition makes a denominator, sign, or domain condition relevant, include that check explicitly.

There is no requirement that every proof use a new quantity. If the desired conclusion follows directly from the hypotheses, a direct proof may be clearer. Introducing an auxiliary quantity is most helpful when it reveals a familiar structure—such as a square, a symmetric product, or a centered expression—that is difficult to see in the original notation.

A good choice can often be found by examining the difference between the two sides of an inequality. If that difference factors into a square or another expression with a known sign, the desired comparison may follow. Alternatively, a substitution can simplify several expressions at once. In either case, the proof should end by stating the original conclusion, not merely a fact about the auxiliary symbol.

Check Your Understanding

For each question, identify what the auxiliary quantity is and how its definition helps establish the original claim.

In the proof that $x^2+y^2\geq2xy$, what quantity is introduced, and which property of it supplies the inequality?
In the identity involving $(x+3)^2$, why is $u=x+3$ a useful choice? Write $x+1$ and $x+5$ in terms of $u$.
Why are the hypotheses $x>0$ and $y>0$ needed in the proof of $\frac{x}{y}+\frac{y}{x}\geq2$?
If $x+y=10$ and $d=x-y$, what identity expresses $x^2+y^2$ in terms of $x+y$ and $d$?
What additional step is needed to determine the equality case after proving an inequality by adding a nonnegative square?
Why does introducing an auxiliary quantity not permit treating it as an independent variable once the original variables have been fixed?

Choice or step	What it provides	What still needs justification
Set \(d=x-y\)	A real quantity whose square is nonnegative.	Expand \(d^2\) correctly and connect it to the claim.
Set \(u=x+3\)	A shorter name for a repeated expression.	Rewrite every related expression consistently.
Divide by \(xy\)	A normalized expression involving ratios.	Verify \(xy\) is nonzero and has the required sign.
Use a square to prove equality cases	A way to identify when a nonnegative term is zero.	Check both that equality forces the square to vanish and that the resulting condition gives equality.