Inequalities — Mathslive.ie

Section 1 of 10

Number Sets

Before solving any inequality, look for the symbol $x \in \dots$ — it tells you which numbers are allowed.

The four sets

$\mathbb{N}$Natural numbers: $1, 2, 3, 4, \dots$

$\mathbb{Z}$Integers: $\dots, -2, -1, 0, 1, 2, \dots$

$\mathbb{Q}$Rationals: fractions $\tfrac{p}{q}$ (integers + decimals that terminate or repeat)

$\mathbb{R}$Reals: everything on the number line (including $\sqrt{2}, \pi, e, \dots$)

Number-line conventions

Marking the answer

$x \in \mathbb{N}$ or $\mathbb{Z}$ → mark dots at each value (separate points)

$x \in \mathbb{R}$ → draw a solid line (continuous range)

$\leq$ or $\geq$ → filled endpoint (included)

$<$ or $>$ → open endpoint (not included)

Section 2 of 10

Linear Inequalities

Solve linear inequalities the same way as equations — except for one rule:

Must learn — the sign-flip rule

Multiplying or dividing both sides by a negative flips the inequality sign.

$4 < 5 \;\Longrightarrow\; -4 > -5$ (both sides $\times (-1)$, sign flipped)

(i) Solve $2x - 1 \leq 3$, $x \in \mathbb{N}$

$2x \leq 4$

$x \leq 2$

$x \in \mathbb{N}$ ⟹ only natural numbers count.

$x \in \{1, 2\}$

On a number line: filled dots at $1$ and $2$ only.

(ii) Solve $1 - 3x \leq 10$, $x \in \mathbb{Z}$

Two methods — pick whichever you prefer.

Method 1: isolate $x$, sign flip when needed.

$-3x \leq 9$

Divide by $-3$ — flip the sign:

$x \geq -3$

Method 2: move the $x$-term to keep it positive — no sign flip needed.

$1 - 3x \leq 10$

$1 - 10 \leq 3x$

$-9 \leq 3x$

$-3 \leq x$

Same answer: $x \geq -3$.

$x \in \mathbb{Z}$ ⟹ dots at $-3, -2, -1, 0, 1, 2, \dots$ (arrow continues right).

(iii) Same inequality, $x \in \mathbb{R}$

Working is identical. The picture changes:

$x \geq -3$

$x \in \mathbb{R}$ ⟹ solid line from $-3$ rightward, filled dot at $-3$ ($\geq$).

(iv) If the inequality were $x > -3$ with $x \in \mathbb{R}$

Solid line from just right of $-3$ rightward, open circle at $-3$ ($-3$ itself not included).

You try

Solve $5 - 2x < 11$ for $x \in \mathbb{R}$.

Isolate $x$. Decide whether you need to flip the sign.

$5 - 2x < 11$

$-2x < 6$

Divide by $-2$ — flip:

$x > -3$

You try

Solve $3x + 7 \geq 1$ for $x \in \mathbb{Z}$. List the smallest five integer solutions.

Isolate $x$. No negative division here so no sign flip.

$3x \geq -6$

$x \geq -2$

$\{-2, -1, 0, 1, 2\}$

Section 3 of 10

Double Inequalities

Must learn — split it

A double inequality like $a \leq f(x) < b$ is really two inequalities. Split, solve each, then combine.

Solve $-3 \leq 2x - 1 < 3$, $x \in \mathbb{R}$

Split into two:

Left: $-3 \leq 2x - 1$

$-2 \leq 2x$

$-1 \leq x$

Right: $2x - 1 < 3$

$2x < 4$

$x < 2$

Combine — $x$ has to satisfy both:

$-1 \leq x < 2$

Number line: solid line from $-1$ to just before $2$, filled dot at $-1$ ($\geq$), open circle at $2$ ($<$).

You try

Solve $1 \leq 2x - 1 < 3$ for $x \in \mathbb{R}$.

Split. Solve each side. Combine.

Left: $1 \leq 2x - 1 \;\Rightarrow\; 2 \leq 2x \;\Rightarrow\; 1 \leq x$

Right: $2x - 1 < 3 \;\Rightarrow\; 2x < 4 \;\Rightarrow\; x < 2$

$1 \leq x < 2$

Section 4 of 10

Quadratic Inequalities

For $ax^{2} + bx + c \;\;\square\;\; 0$ (where $\square$ is one of $<, \leq, >, \geq$): solve the equation first, then look at the parabola.

The temperature analogy

Think of $y = x^{2} - 2x - 8$ where $x$ is time and $y$ is temperature. "$y \leq 0$" asks: when is the temperature at or below zero?

(i) Solve $x^{2} - 2x - 8 \leq 0$

Step 1: solve the equation $x^{2} - 2x - 8 = 0$.

$(x-4)(x+2) = 0$

$x = 4 \;$ or $\; x = -2$

Step 2: sketch the parabola. The $x^{2}$ coefficient is positive, so it opens upward — like a smile. It crosses the $x$-axis at $-2$ and $4$.

Step 3: read the inequality. $\leq 0$ means we want where the curve is at or below the $x$-axis — between the roots.

$-2 \leq x \leq 4$

Must learn — the parabola rule

For $x^{2} + \dots$ (positive $x^{2}$ ⟹ smile-shape):

$\;\;\bullet\;\;\leq 0 \;\;$ or $\;\; < 0$ ⟹ between the roots: $\;\text{smaller} \leq x \leq \text{bigger}$

$\;\;\bullet\;\;\geq 0 \;\;$ or $\;\; > 0$ ⟹ outside the roots: $\;x \leq \text{smaller}$ or $x \geq \text{bigger}$

(ii) Solve $x^{2} - 3x - 10 > 0$

$x^{2} - 3x - 10 = 0$

$(x-5)(x+2) = 0$

$x = 5 \;$ or $\; x = -2$

$> 0$ ⟹ outside the roots:

$x < -2 \;$ or $\; x > 5$

(iii) Solve $9 - x^{2} \geq 0$

Rearrange so $x^{2}$ is positive — then the parabola rule applies cleanly.

$9 - x^{2} \geq 0$

$x^{2} - 9 \leq 0 \;$ (flipped sign)

$(x-3)(x+3) = 0$

$x = 3 \;$ or $\; x = -3$

$\leq 0$ ⟹ between the roots:

$-3 \leq x \leq 3$

(iv) Solve $3x - 2x^{2} < 0$

$3x - 2x^{2} < 0$

$2x^{2} - 3x > 0 \;$ (flipped)

$x(2x - 3) = 0$

$x = 0 \;$ or $\; x = \tfrac{3}{2}$

$> 0$ ⟹ outside the roots:

$x < 0 \;$ or $\; x > \tfrac{3}{2}$

You try

Solve $x^{2} - 5x + 6 \leq 0$.

Factor. Smile-parabola. $\leq 0$ ⟹ between the roots.

$(x-2)(x-3) = 0 \;\Rightarrow\; x = 2 \;$ or $\; x = 3$

$\leq 0$ ⟹ between the roots

$2 \leq x \leq 3$

You try

Solve $4 - x^{2} > 0$.

Rearrange so $x^{2}$ is positive — sign flips.

$4 - x^{2} > 0 \;\Rightarrow\; x^{2} - 4 < 0$

$(x-2)(x+2) = 0 \;\Rightarrow\; x = \pm 2$

$< 0$ ⟹ between the roots

$-2 < x < 2$

Section 5 of 10

Absolute Value Inequalities

$|x|$ means "the distance of $x$ from zero" — always positive. We have two methods.

Method 1 — square both sides

$|a|^{2} = a^{2}$ always. Squaring kills the absolute value bars and gives a quadratic inequality.

Method 2 — split into two cases

$|f(x)| < k$ ⟺ $f(x) < k$ and $f(x) > -k$ (or equivalently $-k < f(x) < k$)

(i) Solve $|x - 1| < 3$ — by squaring

$(x - 1)^{2} < 3^{2}$

$x^{2} - 2x + 1 < 9$

$x^{2} - 2x - 8 < 0$

$(x-4)(x+2) = 0 \;\Rightarrow\; x = 4 \;$ or $\; x = -2$

$< 0$ ⟹ between the roots:

$-2 < x < 4$

(ii) Solve $|2x - 1| < 1$ — by squaring

$(2x-1)^{2} < 1^{2}$

$4x^{2} - 4x + 1 < 1$

$4x^{2} - 4x < 0$

$x^{2} - x < 0$

$x(x-1) = 0 \;\Rightarrow\; x = 0 \;$ or $\; x = 1$

$0 < x < 1$

(iii) Same inequality by splitting

$|2x-1| < 1$ becomes two pieces:

Case A: $2x - 1 < 1 \;\Rightarrow\; 2x < 2 \;\Rightarrow\; x < 1$

Case B: $-(2x - 1) < 1 \;\Rightarrow\; 2x - 1 > -1 \;\Rightarrow\; 2x > 0 \;\Rightarrow\; x > 0$

Combine — needs to be both:

$0 < x < 1 \;\;$ (same answer ✓)

You try

Solve $|x + 2| \leq 5$.

Square both sides — gives a quadratic to factorise.

$(x+2)^{2} \leq 25$

$x^{2} + 4x + 4 \leq 25$

$x^{2} + 4x - 21 \leq 0$

$(x+7)(x-3) = 0 \;\Rightarrow\; x = -7 \;$ or $\; x = 3$

$-7 \leq x \leq 3$

You try

Solve $|3x - 2| < 4$.

Square or split — your choice.

Splitting: $-4 < 3x - 2 < 4$

$-2 < 3x < 6$

$-\tfrac{2}{3} < x < 2$

Section 6 of 10

Fractional Inequalities

When $x$ is in the denominator, you cannot multiply both sides by the bottom directly — you don't know its sign.

Must learn — the trick

Multiply both sides by the bottom squared. Squares are always positive — no sign-flip risk.

Warm-up — positive bottom (constant)

Solve $3 \cdot \dfrac{2x-1}{3} < 1 \cdot 3$.

$2x - 1 < 3$

$x < 2$

If you multiplied by $-3$ instead — flip the sign.

$-3 \cdot \dfrac{2x-1}{-3} < 1 \cdot (-3)$

$2x - 1 > -3$ (flipped)

$2x > -2$

$x > -1$

(i) Solve $\dfrac{x}{x-1} \leq 2$, $x \in \mathbb{R}, \; x \neq 1$

Step 1: multiply both sides by $(x-1)^{2}$.

$\dfrac{x}{x-1} \cdot (x-1)^{2} \leq 2(x-1)^{2}$

$x(x-1) \leq 2(x-1)^{2}$

$x^{2} - x \leq 2(x^{2} - 2x + 1)$

$x^{2} - x \leq 2x^{2} - 4x + 2$

$-x^{2} + 3x - 2 \leq 0$

Multiply by $-1$ (flip the sign):

$x^{2} - 3x + 2 \geq 0$

$(x-1)(x-2) = 0 \;\Rightarrow\; x = 1 \;$ or $\; x = 2$

$\geq 0$ ⟹ outside the roots:

$x \leq 1 \;$ or $\; x \geq 2$

But $x \neq 1$ (the original denominator must not be zero):

$x < 1 \;$ or $\; x \geq 2$

(ii) Solve $\dfrac{x}{2x-1} \leq 1$, $x \neq \tfrac{1}{2}$

$\dfrac{x}{2x-1} (2x-1)^{2} \leq 1 \cdot (2x-1)^{2}$

$x(2x-1) \leq (2x-1)^{2}$

$2x^{2} - x \leq 4x^{2} - 4x + 1$

$-2x^{2} + 3x - 1 \leq 0$

$2x^{2} - 3x + 1 \geq 0 \;$ (flipped)

$(2x-1)(x-1) = 0 \;\Rightarrow\; x = \tfrac{1}{2} \;$ or $\; x = 1$

$\geq 0$ ⟹ outside the roots:

$x \leq \tfrac{1}{2} \;$ or $\; x \geq 1$

But $x \neq \tfrac{1}{2}$:

$x < \tfrac{1}{2} \;$ or $\; x \geq 1$

You try

Solve $\dfrac{x+1}{x-2} \geq 1$, $x \neq 2$.

Multiply by $(x-2)^{2}$. Then expand, rearrange, factor.

$(x+1)(x-2) \geq (x-2)^{2}$

$x^{2} - x - 2 \geq x^{2} - 4x + 4$

$3x \geq 6$

$x \geq 2$

But $x \neq 2$:

$x > 2$

Section 7 of 10

Inequality Proofs — Perfect Squares

Must learn — the key fact

$(\text{any real number})^{2} \geq 0$

Every inequality proof relies on this. Get the LHS into the form (perfect square) $\geq 0$ and you're done.

The technique

Strategy

1. Bring everything to one side. Aim for $0$ on the right.

2. Show the left side is a perfect square (or sum of perfect squares).

3. Conclude: since any number squared is $\geq 0$, LHS $\geq 0$, so the original inequality holds.

Prove $x^{2} + 4 \geq 4x$

Bring everything to one side:

$x^{2} - 4x + 4 \geq 0$

The LHS is a perfect square:

$(x - 2)^{2} \geq 0 \quad\checkmark$

Any real number squared is $\geq 0$ — true for all $x$. Proved.

Prove $a^{2} - ab + b^{2} \geq ab$

$a^{2} - 2ab + b^{2} \geq 0$

$(a - b)^{2} \geq 0 \quad\checkmark$

Any real number squared is $\geq 0$. Proved for all real $a, b$.

Prove $p^{2} + 3q^{2} \geq 2pq$

A bit trickier — only $p^{2} - 2pq + q^{2} = (p-q)^{2}$. We have an extra $2q^{2}$ to spare.

$p^{2} - 2pq + 3q^{2} \geq 0$

Split $3q^{2}$ as $q^{2} + 2q^{2}$:

$\left(p^{2} - 2pq + q^{2}\right) + 2q^{2} \geq 0$

$(p - q)^{2} + 2q^{2} \geq 0 \quad\checkmark$

Sum of two squares — definitely $\geq 0$. Proved.

You try

Prove $x^{2} + 9 \geq 6x$ for all $x \in \mathbb{R}$.

Everything to one side. Look for a perfect square.

$x^{2} - 6x + 9 \geq 0$

$(x - 3)^{2} \geq 0$

Any number squared is $\geq 0$. Proved.

$(x-3)^{2} \geq 0$ — proved.

You try

Prove $a^{2} + b^{2} \geq 2ab$ for all real $a, b$.

Bring $2ab$ over. What perfect square has $a^{2} - 2ab + b^{2}$?

$a^{2} - 2ab + b^{2} \geq 0$

$(a - b)^{2} \geq 0$

True for all real $a, b$. Proved.

$(a-b)^{2} \geq 0$ — proved.

You try

Prove $x^{2} + 2y^{2} \geq 2xy$ for all real $x, y$.

$2y^{2} = y^{2} + y^{2}$ — split it to leave one $y^{2}$ free.

$x^{2} - 2xy + 2y^{2} \geq 0$

$x^{2} - 2xy + y^{2} + y^{2} \geq 0$

$(x - y)^{2} + y^{2} \geq 0$

Sum of squares — proved.

$(x-y)^{2} + y^{2} \geq 0$ — proved.

Section 8 of 10

Harder Proofs — Clearing the Fractions

Some proofs have fractions. The plan: clear the fractions, factor, then reduce to "perfect square $\geq 0$".

Strategy for fraction proofs

1. Bring everything to one side.

2. Combine into a single fraction.

3. Identify the sign of the denominator (often a square — always positive).

4. Show the numerator is $\geq 0$ using a perfect-square argument.

Useful factorisations

Keep these handy

$a^{2} - b^{2} = (a-b)(a+b)$

$a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$

$a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$

Worked — show $\dfrac{1}{a} + \dfrac{1}{b} \geq \dfrac{4}{a+b}$ for $a, b > 0$

Combine the LHS into one fraction:

$\dfrac{b + a}{ab} \geq \dfrac{4}{a+b}$

Cross-multiply (both denominators positive, so safe):

$(a+b)^{2} \geq 4ab$

$a^{2} + 2ab + b^{2} \geq 4ab$

$a^{2} - 2ab + b^{2} \geq 0$

$(a - b)^{2} \geq 0 \quad\checkmark$

True for all real $a, b$ — and in particular for our $a, b > 0$. Proved.

You try

For $x, y > 0$, prove $\dfrac{x}{y} + \dfrac{y}{x} \geq 2$.

Combine over $xy$ (positive). Move the $2$ across. Look for a perfect square in the numerator.

$\dfrac{x^{2} + y^{2}}{xy} \geq 2$

$x^{2} + y^{2} \geq 2xy$ ($xy > 0$, safe to multiply)

$x^{2} - 2xy + y^{2} \geq 0$

$(x - y)^{2} \geq 0$

True for all real $x, y$. Proved.

Reduces to $(x-y)^{2} \geq 0$ — proved.

Section 9 of 10

State Exam Standard Proof

Let $a, b$ be non-zero reals with $a + b \geq 0$. Show that $\dfrac{a}{b^{2}} + \dfrac{b}{a^{2}} \geq \dfrac{1}{a} + \dfrac{1}{b}$.

Step 1: combine each side into a single fraction.

LHS: $\dfrac{a}{b^{2}} + \dfrac{b}{a^{2}} = \dfrac{a^{3} + b^{3}}{a^{2}b^{2}}$

RHS: $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{a + b}{ab}$

Step 2: the inequality becomes

$\dfrac{a^{3} + b^{3}}{a^{2}b^{2}} \geq \dfrac{a + b}{ab}$

Cross-multiply by $a^{2}b^{2}$ (it's a square — always positive):

$a^{3} + b^{3} \geq \dfrac{(a+b) \cdot a^{2}b^{2}}{ab} = ab(a + b)$

$a^{3} + b^{3} \geq a^{2}b + ab^{2}$

Step 3: factor each side using $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$.

$(a + b)(a^{2} - ab + b^{2}) \geq ab(a + b)$

Step 4: $a + b \geq 0$ given, so we can divide both sides by $(a + b)$ without flipping (careful: if $a + b = 0$ both sides are $0$, equality holds):

$a^{2} - ab + b^{2} \geq ab$

$a^{2} - 2ab + b^{2} \geq 0$

$(a - b)^{2} \geq 0 \quad\checkmark$

Any real squared is $\geq 0$. Proved.

Section 10 of 10

Worded Inequalities

Common phrasing → symbols:

Translating words

"at most" or "no more than" → $\leq$

"at least" or "no less than" → $\geq$

"less than" → $<$

"more than" or "greater than" → $>$

"between $a$ and $b$" → $a < x < b$ (or $\leq$ if "inclusive")

Quick translations

"Maths grade between $11$ and $12$ inclusive" → $11 \leq M \leq 12$

"Tractor at least $16$" → $T \geq 16$

"Work hours between $16$ and $66$ inclusive" → $16 \leq W \leq 66$

Worded — daily wage

A worker earns at least €70 per day. They work no more than $8$ hours. Express the hourly rate $r$ as an inequality.

$8 r \geq 70$ ⟹ $r \geq \tfrac{70}{8}$

$r \geq 8.75$ (€ per hour)

You try

A car park charges €2 entry plus €1.50 per hour. You have €11 to spend. How many full hours $h$ can you park?

Total cost is $2 + 1.5h$. Spend "at most" $11$ ⟹ $\leq$. Then $h \in \mathbb{N}$ for full hours.

$2 + 1.5h \leq 11$

$1.5h \leq 9$

$h \leq 6$

Up to $6$ full hours.

$h \leq 6$ ⟹ up to 6 hours

You try

A test has $20$ questions, each worth $5$ marks. You need at least $70\%$. What is the minimum number of correct answers?

Total marks $= 100$. $70\% = 70$. Each correct $= 5$ marks. "At least" ⟹ $\geq$.

$5n \geq 70$

$n \geq 14$

At least $14$ correct answers.

At least $14$

That's Inequalities.

You've covered linear, double, quadratic, absolute value, fractional, and proofs — including the State Exam style with $a + b \geq 0$.