Grade 10

Inequalities

Statements comparing two expressions with <, ≤, > or ≥, including classical bounds like AM–GM.

IntuitionComparing Sizes Without Computing Exactly

Not every mathematical question asks "what is the exact value?" — often we only need to know which of two quantities is larger, or that one never exceeds the other. The symbols $a<b$ , $a\le b$ , $a>b$ , $a\ge b$ record exactly this kind of comparison: for instance $3<5$ is obviously true, and $x^2\ge 0$ holds no matter what real value we substitute. Inequalities let us reason about an entire family of numbers at once — exactly what is needed to bound a measurement error, optimize the shape of a container, or guarantee that an algorithm never runs too slowly.

The curve

f(x)=x^3-3x

: the regions where the curve sits above or below the horizontal axis are exactly the solution sets of the inequalities

f(x)\ge 0

and

f(x)\le 0

. Drag

a

c

to see how the sign pattern (and hence the solution set) changes.

SchoolRules for Manipulating Inequalities

Definition: Order relations

For real numbers $a$ and $b$ , we write $a<b$ when $b-a$ is positive, and $a\le b$ when $b-a$ is nonnegative; $a>b$ and $a\ge b$ are defined symmetrically. An inequality is strict if it uses $<$ or $>$ , and non-strict if it uses $\le$ or $\ge$ .

a<b \;\Longleftrightarrow\; a+c<b+c \quad \text{for every real } c

Adding the same number to both sides never changes the direction of an inequality: $a<b \Rightarrow a+c<b+c$ . Multiplying by a positive number also preserves the direction, $a<b,\ c>0 \Rightarrow ac<bc$ , but multiplying by a negative number reverses it, $a<b,\ c<0 \Rightarrow ac>bc$ — this sign flip is the single most common source of errors when solving inequalities.

a<b,\ c<0 \;\Longrightarrow\; ac>bc

Effect of an operation on both sides of $a<b$
Operation	Result
Add $c$ (any real)	Direction unchanged: $a+c<b+c$
Multiply by $c>0$	Direction unchanged: $ac<bc$
Multiply by $c<0$	Direction reverses: $ac>bc$
Square, if $a,b\ge 0$	Direction unchanged: $a^2<b^2$

UndergraduateTwo Classical Inequalities

Theorem: AM–GM inequality (two numbers)

For all real numbers $a\ge 0$ and $b\ge 0$ , $\dfrac{a+b}{2} \ge \sqrt{ab}$ , with equality exactly when $a=b$ .

Why is it true?

The arithmetic mean $\frac{a+b}{2}$ treats "spread out" numbers gently, while the geometric mean $\sqrt{ab}$ punishes spread: multiplying two very different numbers gives a smaller "typical size" than adding and halving them.

Proof

Start from a fact that is always true: any real square is nonnegative. Apply this to $\sqrt{a}-\sqrt{b}$ , which is a real number since $a,b\ge 0$ : $(\sqrt{a}-\sqrt{b})^2 \ge 0$ .

Expand the square using $(u-v)^2=u^2-2uv+v^2$ with $u=\sqrt a$ , $v=\sqrt b$ : this gives $a - 2\sqrt{ab} + b \ge 0$ , i.e. $a - 2\sqrt{ab} + b \ge 0$ .

Rearrange by adding $2\sqrt{ab}$ to both sides and then dividing everything by $2$ (a positive number, so the direction is preserved): $\dfrac{a+b}{2} - \sqrt{ab} \ge 0$ , which is exactly $\dfrac{a+b}{2} \ge \sqrt{ab}$ .

Equality holds throughout only when the very first step was equality, i.e. $(\sqrt a-\sqrt b)^2=0$ , which happens exactly when $\sqrt a=\sqrt b$ , i.e. $a=b$ .

Theorem: Triangle inequality (for real numbers)

For all real numbers $a$ and $b$ , $|a+b| \le |a|+|b|$ , where $|x|$ denotes absolute value; equality holds exactly when $a$ and $b$ have the same sign (or one is $0$ ).

Why is it true?

Absolute value measures distance from zero; adding two numbers of opposite sign causes cancellation, so the distance of the sum can only shrink compared to adding the distances separately.

Proof

Every real number $x$ satisfies $-|x|\le x\le |x|$ by definition of absolute value. Apply this to both $a$ and $b$ : $-|a|\le a\le |a|$ and $-|b|\le b\le |b|$ .

Add the two chains of inequalities term by term (allowed since adding preserves direction): $-(|a|+|b|)\le a+b\le |a|+|b|$ .

The statement $-(M)\le y\le M$ for $M\ge 0$ is exactly equivalent to $|y|\le M$ by definition of absolute value; here $y=a+b$ and $M=|a|+|b|$ , so $|a+b|\le |a|+|b|$ , which is $|a+b| \le |a|+|b|$ .

Equality requires both chained inequalities to be equalities simultaneously, which forces $a$ and $b$ to have the same sign (both nonnegative or both nonpositive), since only then does no cancellation occur between $a$ and $b$ .

UndergraduateReal-World Applications and Worked Examples

Inequalities are the everyday tool of engineers and scientists who never need an exact answer, only a guaranteed bound: a bridge must hold at least a given load, a signal-to-noise ratio must exceed a threshold, a budget must not exceed a cap. The AM–GM inequality in particular turns "what is the best possible shape or allocation?" into pure algebra, because a sum-with-fixed-product or product-with-fixed-sum problem is exactly what $\dfrac{a+b}{2} \ge \sqrt{ab}$ controls.

Example: Maximizing the area of a fenced garden

A gardener has $40$ meters of fencing to enclose a rectangular garden of width $x$ and length $y$ , so $2x+2y=40$ . What is the largest possible area $S=xy$ , and for which $x,y$ is it achieved?

Solution

From $2x+2y=40$ we get $x+y=20$ . We want to maximize $S=xy$ subject to this fixed sum.

By the AM–GM inequality, $\dfrac{a+b}{2} \ge \sqrt{ab}$ applied to $x,y\ge 0$ gives $10=\frac{x+y}{2}\ge\sqrt{xy}=\sqrt{S}$ , so $\sqrt S\le 10$ , hence $S\le 100\text{ m}^2$ .

Equality in AM–GM holds exactly when $x=y$ , so $x=y=10$ — a square garden — achieves the bound, giving $S=10\times 10=100\text{ m}^2$ .

Any other split, e.g. $x=5,y=15$ , gives only $S=75\text{ m}^2 < 100\text{ m}^2$ , confirming the square is optimal.

Example: Bounding total GPS position error

A GPS receiver combines a horizontal error of $a=3$ meters from atmospheric delay with a further error of $b=-2$ meters from clock drift (a negative value here means it partially cancels the first error). Use the triangle inequality $|a+b| \le |a|+|b|$ to give a guaranteed upper bound on the size of the combined error $|a+b|$ without needing to know the exact sign relationship in general.

Solution

The triangle inequality $|a+b| \le |a|+|b|$ holds for all real $a,b$ , regardless of sign, so it applies here even though we suspect partial cancellation.

Compute the right-hand side using the actual values: $|a|+|b|=|3|+|-2|=3+2=5$ meters. This is the guaranteed worst-case bound: $|a+b|\le 5$ meters, valid even if we did not know the signs of $a,b$ in advance (which is the realistic situation for a receiver designer bounding errors before a specific reading comes in).

In this particular numeric case we can also compute exactly: $a+b=3+(-2)=1$ , so $|a+b|=1$ meter, comfortably inside the bound of $5$ meters — illustrating that the triangle inequality bound is a safe worst case, not always tight, which is exactly what a guaranteed engineering bound should be.

The gap between the bound ( $5$ m) and the actual value ( $1$ m) here reflects that $a$ and $b$ have opposite signs; the bound would be tight (equal to $5$ ) only if both errors pointed the same direction.

For $a=4$ and $b=9$ , what are the arithmetic mean $\frac{a+b}{2}$ and geometric mean $\sqrt{ab}$ ?

Solve the inequality $2x-3>1$ for $x$ .

A gardener has $40$ meters of fencing ( $2x+2y=40$ ) for a rectangular garden of area $S=xy$ . What is the maximum possible area?

If $a<b$ and $c<0$ , which of the following is always true?