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Module 11 · L03 of 10 ~45 min ⚡ +90 XP available

Inequalities with Absolute Value & the Triangle Inequality

Walk from your house to a friend's place via the corner shop, and the route is always at least as long as the direct path. That single geometric truth — the triangle inequality — is the most-quoted lemma in higher mathematics. In Ext 2 you must prove it from the definition of $|x|$ and then deploy it (forward and reverse) to bound expressions, prove convergence claims and tame error estimates.

Today's hook — Let $a = 5$ and $b = -3$. Compute $|a+b|$, $|a|+|b|$, and $\bigl||a|-|b|\bigr|$. Which two are equal here? Now try $a = 5$, $b = 3$ — does the picture change? Compare after card 05.

0/5QUESTS

Recall — your gut answer first

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Without looking anything up, write down a definition of $|x|$ that works for every real number. Then sketch the graph $y = |x|$ on a quick set of axes. Can your definition handle $x = 0$ cleanly?

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The two moves for absolute-value proofs

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Almost every absolute-value inequality proof reduces to squaring (because $|x|^2 = x^2$) or case-splitting on signs. Pick the cleaner of the two before you start writing — that single choice usually decides whether the proof is two lines or two pages.

The square-or-split strategy: (1) if both sides of the inequality are non-negative, square to remove the bars; (2) otherwise, split into cases by sign of each variable; (3) collect terms and identify a sum of squares or a known non-negative quantity to finish.

Definition: $|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}$ · Equivalent: $|x| = \sqrt{x^2}$, so $|x|^2 = x^2$.

$|a+b| \leq |a|+|b|$

$|x|^2 = x^2$ always

The identity $|x|^2 = x^2$ holds for every real $x$, with no case analysis. This single equation is the workhorse of triangle-inequality proofs because squaring removes absolute-value bars without losing information.

$ab \leq |ab| = |a||b|$

For any reals $a, b$, the product $ab$ never exceeds its absolute value, and $|ab| = |a||b|$. This is the only non-obvious step inside the squared form of the triangle inequality.

Reverse triangle

The reverse form $\bigl||a|-|b|\bigr| \leq |a-b|$ is obtained by applying the forward inequality to $a = (a-b)+b$. It bounds the difference of magnitudes by the magnitude of the difference.

What you'll master

Know

Key facts

Definition of $|x|$ and the identity $|x|^2 = x^2$ for all real $x$
Triangle inequality: $|a+b| \leq |a|+|b|$ for all real $a,b$
Reverse triangle inequality: $\bigl||a|-|b|\bigr| \leq |a-b|$

Understand

Concepts

Why squaring both sides is a valid proof move when both sides are non-negative
Why equality in the triangle inequality requires $a$ and $b$ to have the same sign (or one to be zero)
How the reverse triangle inequality follows from the forward one

Can do

Skills

Prove $|a+b| \leq |a|+|b|$ from first principles using $ab \leq |a||b|$
Deduce $|a-b| \geq \bigl||a|-|b|\bigr|$ and apply it to bound expressions
Combine triangle inequality steps to prove multi-term bounds

Key terms

Absolute value $|x|$The non-negative real $|x| = x$ if $x \geq 0$ and $|x| = -x$ if $x < 0$. Equivalently $|x| = \sqrt{x^2}$, so $|x|^2 = x^2$.

Triangle inequalityFor all real $a, b$: $|a + b| \leq |a| + |b|$. Equality holds iff $a$ and $b$ have the same sign (including zero).

Reverse triangle inequalityFor all real $a, b$: $\bigl||a| - |b|\bigr| \leq |a - b|$. The "gap" between magnitudes is bounded by the magnitude of the difference.

Generalised triangle inequalityFor real $a_1, \ldots, a_n$: $\bigl|\sum a_i\bigr| \leq \sum |a_i|$. Proved by induction from the two-term case.

Bounding lemma$-|x| \leq x \leq |x|$ for every real $x$. Adding two such pairs gives $-(|a|+|b|) \leq a+b \leq |a|+|b|$, an alternative route to the triangle inequality.

NESA MEX-P1NESA outcome: proves results using a variety of techniques and considers the validity of an argument — specifically prove and use the triangle inequality and inequalities involving absolute value.

From the definition to the squared form

core concept

Everything you prove about $|x|$ flows from one of two equivalent definitions:

Piecewise: $|x| = x$ if $x \geq 0$ and $|x| = -x$ if $x < 0$.
Radical: $|x| = \sqrt{x^2}$ (using the non-negative square root).

The radical form gives the master identity $|x|^2 = x^2$, which converts every absolute-value question into a polynomial question — provided both sides of the original inequality are non-negative, because squaring is monotonic only on $[0,\infty)$.

Worked through the hook: For $a = 5, b = -3$: $|a + b| = |2| = 2$ and $|a| + |b| = 5 + 3 = 8$, so $|a+b| < |a|+|b|$ (strict). Also $\bigl||a|-|b|\bigr| = |5-3| = 2 = |a+b|$ — equality in the reverse form. Swap to $a = 5, b = 3$: $|a+b| = 8 = |a|+|b|$ — equality in the forward form because $a, b$ have the same sign.

Why squaring works. If $L \geq 0$ and $R \geq 0$ then $L \leq R \iff L^2 \leq R^2$. Both sides of the triangle inequality are absolute values, so both are non-negative, and the squared inequality $|a+b|^2 \leq (|a|+|b|)^2$ is equivalent to the original. After squaring, the bars vanish and you are left with a polynomial inequality reducing to $ab \leq |a||b|$.

Definition: $|x| = x$ for $x \geq 0$, $|x| = -x$ for $x < 0$; equivalently $|x| = \sqrt{x^2}$ · Master identity: $|x|^2 = x^2$ for all real $x$ · Hook: $a=5, b=-3$ gives strict inequality; $a=5, b=3$ gives equality · Squaring is reversible only when both sides are non-negative

Pause — copy the definition $|x| = \sqrt{x^2}$, the master identity $|x|^2 = x^2$, and the forward triangle inequality $|a+b| \leq |a|+|b|$ with its proof strategy into your book.

Quick check: Which identity is valid for every real number $x$ and is the basis of the squared form of the triangle inequality?

The forward and reverse triangle inequalities

core concept

We just saw that $|x|^2 = x^2$ for all real $x$, which lets us square both sides of $|a + b| \leq |a| + |b|$ to prove it algebraically. That raises a question: is there a companion inequality bounding $\bigl||a| - |b|\bigr|$? This card answers it → the reverse triangle inequality $\bigl||a| - |b|\bigr| \leq |a - b|$, proved by applying the forward inequality to $a = (a-b) + b$.

The forward form bounds the magnitude of a sum by the sum of magnitudes. The reverse form bounds how far apart two magnitudes can be by the magnitude of their difference. Both follow from the same key step $ab \leq |a||b|$.

Forward (statement). For all real $a, b$:

$$|a+b| \;\leq\; |a| + |b|$$

Reverse (statement). For all real $a, b$:

$$\bigl||a| - |b|\bigr| \;\leq\; |a - b|$$

Quick deduction of the reverse from the forward: write $a = (a-b) + b$. The forward inequality gives $|a| \leq |a-b| + |b|$, so $|a| - |b| \leq |a-b|$. By symmetry $|b| - |a| \leq |a-b|$, and combining the two yields $\bigl||a|-|b|\bigr| \leq |a-b|$.

Application snapshot. If $|a| \leq 3$ and $|b| \leq 2$ then by the forward inequality $|a + b| \leq 5$. If instead you know $|a-b| \leq 0.1$ then by the reverse inequality $\bigl||a|-|b|\bigr| \leq 0.1$ — useful when proving that a sequence of magnitudes converges as the underlying sequence converges.

Forward: $|a+b| \leq |a|+|b|$ for all real $a,b$ · Reverse: $\bigl||a|-|b|\bigr| \leq |a-b|$ for all real $a,b$ · Trick for reverse: apply forward to $a = (a-b)+b$ then symmetric copy · Use forward to bound sums; use reverse to bound differences of magnitudes

Pause — copy the reverse triangle inequality $\bigl||a|-|b|\bigr| \leq |a-b|$ and the trick of writing $a = (a-b)+b$ then applying the forward inequality into your book.

Did you get this? True or false: for all real $a, b$, $|a - b| \geq \bigl||a| - |b|\bigr|$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · PROVE THE TRIANGLE INEQUALITY

Prove that for all real $a, b$: $|a + b| \leq |a| + |b|$.

Both $|a+b|$ and $|a|+|b|$ are non-negative, so it suffices to prove the squared inequality $|a+b|^2 \leq (|a|+|b|)^2$. Use $|x|^2 = x^2$:
$|a+b|^2 = (a+b)^2 = a^2 + 2ab + b^2$.

Square both sides — legitimate because both sides are $\geq 0$, and squaring removes the bars cleanly via the master identity.

PROBLEM 2 · APPLYING THE TRIANGLE INEQUALITY TO BOUND

Given $|x - 2| < 0.1$ and $|y + 3| < 0.2$, prove that $|x + y - (-1)| = |x + y + 1| < 0.3$.

Rewrite the target as a sum of the two given quantities:
$x + y + 1 = (x - 2) + (y + 3)$.

The trick in every "bound this" question is to express the target as a sum (or scalar multiple) of quantities that already have known bounds.

PROBLEM 3 · REVERSE TRIANGLE INEQUALITY

Prove that for all real $a, b$: $\bigl||a| - |b|\bigr| \leq |a - b|$.

Write $a = (a - b) + b$ and apply the forward triangle inequality to $a-b$ and $b$:
$|a| = |(a-b) + b| \leq |a-b| + |b|$.
Rearranging: $|a| - |b| \leq |a-b|$.

Use the forward inequality already proven. The rearrangement is valid because we are subtracting a real number from both sides.

Fill the gap: The squared form of the triangle inequality reduces, after expansion, to the inequality $2ab \leq 2|a||b|$, which holds because for any real numbers $a$ and $b$ we have $ab \leq |ab| = |a|\cdot|b|$. Equality in the original triangle inequality therefore requires the product $ab$ to be .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Writing $|x| = x$ as if it were universal

The definition is piecewise: $|x| = x$ only when $x \geq 0$. Writing "let $|x| = x$" with no condition forces $x \geq 0$ and silently throws away half the cases. Either use the piecewise form, or use the radical form $|x| = \sqrt{x^2}$ which holds for every real $x$ without splitting.

Trap 02

Squaring when one side might be negative

Squaring is monotonic on $[0,\infty)$ only. If you start from an inequality where one side could be negative — for example, $a + b \leq |a| + |b|$ rather than $|a+b| \leq |a|+|b|$ — squaring can reverse the direction. Always confirm both sides are non-negative before you square.

Trap 03

Confusing $|a - b|$ with $|a| - |b|$

$|a - b|$ is the distance between $a$ and $b$; it is always $\geq 0$. $|a| - |b|$ can be negative (try $a=1, b=3$: $|a|-|b| = -2$ while $|a-b| = 2$). The reverse triangle inequality $\bigl||a|-|b|\bigr| \leq |a-b|$ corrects for this — note the outer bars on the left.

Did you get this? True or false: equality holds in $|a+b| \leq |a|+|b|$ if and only if $a$ and $b$ have the same sign (where either being zero is allowed).

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Starting from the definition $|x| = \sqrt{x^2}$, prove that $|xy| = |x||y|$ for all real $x, y$.

Given $|x - 5| < 0.05$ and $|y - 2| < 0.03$, prove $|(x+y) - 7| < 0.08$. State the inequality you use at each step.

Use the reverse triangle inequality to prove: if $|x_n - L| \to 0$ then $\bigl||x_n| - |L|\bigr| \to 0$.

Prove that for all real $a, b, c$: $|a + b + c| \leq |a| + |b| + |c|$. (Apply the two-term inequality twice.)

Determine all real $a, b$ for which equality holds in $|a + b| = |a| + |b|$. Justify using the proof of the inequality.

Odd one out: Three of these statements hold for all real $a, b$. Which one is NOT a valid consequence of the triangle inequality?

Revisit your thinking

Earlier you wrote down your own definition of $|x|$ and tried the hook with $a = 5, b = \pm 3$.

The piecewise definition is the one most students reach for, but the radical form $|x| = \sqrt{x^2}$ is what makes the triangle-inequality proof short: squaring is just removing a square root. With $a = 5, b = -3$ you get strict inequality (opposite signs); with $a = 5, b = 3$ equality holds because the two share a sign. The reverse form bites whenever the signs disagree.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Use the triangle inequality to prove that for all real $x, y$, $|x - y| \leq |x| + |y|$. (2 marks)

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ApplyBand 43 marks

Q2. Prove from the definition that $|a + b| \leq |a| + |b|$ for all real $a, b$. State clearly where equality occurs. (3 marks)

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AnalyseBand 54 marks

Q3. Suppose $|x - a| < \tfrac{\varepsilon}{2}$ and $|y - b| < \tfrac{\varepsilon}{2}$ where $\varepsilon > 0$. Prove that $|(x + y) - (a + b)| < \varepsilon$. Then use the reverse triangle inequality to show $\bigl||x| - |a|\bigr| < \tfrac{\varepsilon}{2}$. (4 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $|xy| = \sqrt{(xy)^2} = \sqrt{x^2 y^2} = \sqrt{x^2}\sqrt{y^2} = |x||y|$. Each step uses non-negativity of the square root and $\sqrt{ab} = \sqrt{a}\sqrt{b}$ for $a,b \geq 0$.

2. $(x+y) - 7 = (x-5)+(y-2)$. By the triangle inequality $|(x+y)-7| \leq |x-5| + |y-2| < 0.05 + 0.03 = 0.08$.

3. By the reverse triangle inequality $0 \leq \bigl||x_n|-|L|\bigr| \leq |x_n - L| \to 0$. By the squeeze, $\bigl||x_n|-|L|\bigr| \to 0$.

4. $|a+b+c| = |(a+b)+c| \leq |a+b|+|c| \leq (|a|+|b|)+|c| = |a|+|b|+|c|$. Triangle inequality applied twice.

5. Equality in the proof requires $2ab = 2|a||b|$, i.e. $ab \geq 0$. So $a, b$ have the same sign (one being zero is allowed). Geometrically the "two displacements" point the same way.

Q1 (2 marks): $x - y = x + (-y)$ [1]. By the triangle inequality $|x + (-y)| \leq |x| + |-y| = |x| + |y|$ [1].

Q2 (3 marks): Both sides non-negative so square: $|a+b|^2 = (a+b)^2 = a^2 + 2ab + b^2$ [1]. $(|a|+|b|)^2 = a^2 + 2|a||b| + b^2$ [1]. Since $ab \leq |a||b|$, the squared inequality holds; take square roots. Equality iff $ab = |a||b|$, i.e. $a, b$ share a sign [1].

Q3 (4 marks): $(x+y)-(a+b) = (x-a)+(y-b)$; triangle inequality gives $|(x+y)-(a+b)| \leq |x-a|+|y-b| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon$ [2]. Reverse triangle inequality: $\bigl||x|-|a|\bigr| \leq |x - a| < \tfrac{\varepsilon}{2}$ [2].

Boss battle · The Inequality Forge

earn bronze · silver · gold

Five timed questions on the triangle inequality and its reverse — proofs, bounding, and equality conditions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering absolute-value and triangle-inequality questions. Lighter alternative to the boss.

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Module overview · Extension 2