Trivial Inequality

The trivial inequality states that for any real number x, the square x² is always greater than or equal to zero. Written formally: x² ≥ 0 for all x ∈ ℝ, with equality holding if and only if x = 0. Although this statement may seem obvious — after all, squaring any number always produces a non-negative result — it turns out to be one of the most powerful and frequently used tools in mathematical proof-writing, especially in competition mathematics and analysis. The reason mathematicians call it 'trivial' is not that it is unimportant, but that its proof requires almost no work: a positive number squared is positive, a negative number squared is positive (since a negative times a negative is positive), and zero squared is zero. Despite its simple proof, the inequality unlocks an enormous range of results. By clever substitution or algebraic manipulation, many non-obvious inequalities collapse into a direct application of x² ≥ 0. In the context of linear inequalities, the trivial inequality bridges algebra and proof. You will use it to prove results like the AM-GM inequality, to establish bounds on expressions, and to verify that certain quadratic expressions can never be negative. Understanding this concept deeply will sharpen your algebraic intuition and prepare you for more advanced mathematics, including calculus, number theory, and olympiad problem solving.

Key takeaways
  • Statement and Proof
  • Using Substitution to Prove Other Inequalities
  • Equality Conditions Matter
  • Combining Multiple Applications
  • Connection to Completing the Square

Statement and Proof

For any real number x, we have x² ≥ 0. Proof: If x > 0, then x² = x · x > 0. If x < 0, then x² = x · x = (negative)(negative) > 0. If x = 0, then x² = 0. In all cases x² ≥ 0, and x² = 0 exactly when x = 0. This completes the proof. The key insight is that the sign of x does not matter — squaring always removes it.

Using Substitution to Prove Other Inequalities

The real power comes from substituting expressions for x. For instance, replacing x with (a − b) gives (a − b)² ≥ 0, which expands to a² − 2ab + b² ≥ 0, and therefore a² + b² ≥ 2ab. This is the algebraic AM-GM inequality for two terms: (a² + b²)/2 ≥ ab when a, b ≥ 0 (set a → √a, b → √b). A single line of algebra, anchored by the trivial inequality, proves a result that took mathematicians centuries to generalise.

Equality Conditions Matter

Whenever you apply the trivial inequality, you must track when equality holds. Since x² = 0 iff x = 0, equality in (a − b)² ≥ 0 holds iff a = b. This equality condition is just as important as the inequality itself — it tells you exactly when the bound is tight and often pinpoints the maximum or minimum of an expression.

Combining Multiple Applications

You can add several instances of the trivial inequality together. For example: (a − b)² ≥ 0 and (b − c)² ≥ 0 and (a − c)² ≥ 0. Adding all three: 2a² + 2b² + 2c² − 2ab − 2bc − 2ac ≥ 0, so a² + b² + c² ≥ ab + bc + ca. This technique — summing non-negative squares — is a cornerstone strategy in inequality proofs.

Connection to Completing the Square

Completing the square is essentially the trivial inequality in disguise. Any quadratic ax² + bx + c (with a > 0) can be written as a(x + b/(2a))² + (c − b²/(4a)). Since the squared term is ≥ 0, the minimum value of the quadratic is c − b²/(4a), achieved at x = −b/(2a). The trivial inequality thus gives you the vertex of a parabola without calculus.

The Key Equation

x2≥0for all x∈R, with equality iff x=0x^2 \geq 0 \quad \text{for all } x \in \mathbb{R}, \text{ with equality iff } x = 0
  • x — Real number (dimensionless): Any element of the real number line, positive, negative, or zero.
  • x^2 — Square of x (dimensionless): The product x · x; always non-negative regardless of the sign of x.
  • \geq — Greater than or equal to (—): The inequality sign indicating the left side is at least as large as the right side.
  • 0 — Zero (dimensionless): The lower bound; the smallest value that x² can attain.

Worked Example

Problem. Prove that for all real numbers a and b, the inequality a² + b² + 1 ≥ ab + a + b holds.

(a−b)2+(a−1)2+(b−1)2≥0  ⟹  a2+b2+1≥ab+a+b(a-b)^2 + (a-1)^2 + (b-1)^2 \geq 0 \implies a^2+b^2+1 \geq ab+a+b

Answer. The inequality holds for all real a, b, with equality exactly when a = b = 1.

Figure
Diagram
Graph of y = x² (always ≥ 0)xy(0, 0)y = 0y = x²-3-2-1012312468Minimum at x=0, y=0Parabola lies entirelyon or above y = 0
Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · French

Cauchy was a pioneer of rigorous analysis who formalised many fundamental inequalities, including the Cauchy–Schwarz inequality. The proof of the Cauchy–Schwarz inequality — that (Σaᵢbᵢ)² ≤ (Σaᵢ²)(Σbᵢ²) — ultimately rests on the trivial inequality applied to a discriminant argument. Cauchy's insistence on rigorous proof rather than intuitive reasoning transformed mathematics, and his work on inequalities set the template for how the trivial inequality is used in modern analysis. Cauchy published an astonishing 789 papers over his lifetime — so many that the journal Comptes Rendus had to introduce a page limit specifically to manage his prolific submissions.

  • AM-GM Inequality — The two-variable AM-GM inequality (a+b)/2 ≥ √(ab) is proved directly by applying the trivial inequality to (√a − √b)² ≥ 0, making it the most immediate application of the concept.
  • Completing the Square — Completing the square rewrites a quadratic as a sum involving a perfect square plus a constant, using the trivial inequality to determine the minimum value of the quadratic expression.
  • Cauchy–Schwarz Inequality — The Cauchy–Schwarz inequality can be proved by considering a quadratic in a real parameter t and applying the trivial inequality to show its discriminant must be non-positive.
  • Absolute Value Inequalities — The absolute value |x| equals √(x²), so properties of absolute values are closely linked to the trivial inequality and the non-negativity of squares.