Prove and apply inequalities including the use of the discriminant, completing the square, and standard results such as the AM-GM inequality

What this dot point is asking

SEAB wants you to prove inequalities rigorously and to solve inequalities that arise in algebra. You should be fluent with the "consider the difference" method, completing the square, the discriminant condition for a quadratic, and standard results such as the arithmetic mean-geometric mean (AM-GM) inequality. The recurring idea is that a real square is never negative, which underlies most proofs.

The answer

The fundamental fact

The whole topic rests on one observation:

Most inequality proofs reduce to showing that some expression is a sum of squares.

Proving an inequality: consider the difference

To prove $A \geq B$, examine the difference $A - B$ and show it is non-negative, typically by writing it as a square or a sum of squares. For example $a^2 + b^2 \geq 2ab$ follows from $a^2 + b^2 - 2ab = (a-b)^2 \geq 0$.

Completing the square

Any quadratic can be written as

If $a > 0$ the squared term is non-negative, so the minimum value is $c - \dfrac{b^2}{4a}$, attained at $x = -\dfrac{b}{2a}$. This both proves inequalities and locates extrema.

The discriminant condition

For a quadratic $ax^2 + bx + c$ with real coefficients, the discriminant $\Delta = b^2 - 4ac$ controls the roots:

• $\Delta > 0$: two distinct real roots;
• $\Delta = 0$: one repeated real root;
• $\Delta < 0$: no real roots, so the quadratic keeps a constant sign (the sign of $a$).

The case $\Delta < 0$ with $a > 0$ means $ax^2 + bx + c > 0$ for all real $x$, a powerful way to prove an inequality holds everywhere.

Solving polynomial and rational inequalities

To solve an inequality, bring everything to one side so it compares with $0$, factor, and analyse the sign on each interval determined by the critical values. For rational inequalities never multiply across by an expression of unknown sign; instead combine into a single fraction and test signs.

The AM-GM inequality

For non-negative reals, the arithmetic mean is at least the geometric mean. For two terms:

with equality when $a = b$. This follows directly from $(\sqrt{a} - \sqrt{b})^2 \geq 0$.

Examples in context

Example 1. Showing a quadratic is always positive. To prove $2x^2 - 3x + 5 > 0$ for all real $x$, compute $\Delta = 9 - 40 = -31 < 0$; with a positive leading coefficient this guarantees the expression is positive everywhere, a one-line proof.

Example 2. Optimisation by AM-GM. For a positive variable, $x + \dfrac{1}{x} \geq 2$ follows from AM-GM with $a = x$, $b = \dfrac{1}{x}$, giving the minimum value $2$ at $x = 1$. This kind of bound appears throughout optimisation arguments.

Try this

Q1. Prove that $x^2 - 6x + 10 > 0$ for all real $x$. [2 marks]

• Cue. Complete the square: $(x-3)^2 + 1 \geq 1 > 0$, or note $\Delta = 36 - 40 < 0$ with positive leading coefficient.

Q2. Solve $x^2 - x - 6 \leq 0$. [2 marks]

• Cue. Factor $(x-3)(x+2) \leq 0$, which holds between the roots: $-2 \leq x \leq 3$.

Q3. State the AM-GM inequality for two non-negative numbers and the equality condition. [1 mark]

• Cue. $\dfrac{a+b}{2} \geq \sqrt{ab}$ for $a, b \geq 0$, with equality when $a = b$.