Inequalities is the habit of comparing quantities through standard bounds, transformations, and equality cases. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
$$x^2 \ge 0 \text{ for all real } x,\quad\text{with equality iff } x = 0.$$
Most inequalities ultimately rest on the fact that a real square is never negative; the art is rewriting the target as a sum of squares or applying a named inequality, then checking the equality case.
Why It Works
1
A real square is never negative: $(a - b)^2 \ge 0$ for all reals, with equality iff $a = b$.
For positive $a, b$, substitute $a \to \sqrt a,\ b \to \sqrt b$ in $a^2 + b^2 \ge 2ab$ to get $a + b \ge 2\sqrt{ab}$, i.e. $\frac{a + b}{2} \ge \sqrt{ab}$ (AM-GM), equality iff $a = b$.
4
General strategy: to prove $A \ge B$, show $A - B$ is a sum of squares (or a nonnegative combination of these bounds), and always pin down the equality case to confirm the bound is tight.
Worked Examples
Example 1
Prove that $a^2 + b^2 + c^2 \ge ab + bc + ca$ for all real $a, b, c$.
1
Consider $2(a^2 + b^2 + c^2) - 2(ab + bc + ca)$.
2
Group as squares: this equals $(a - b)^2 + (b - c)^2 + (c - a)^2$.
3
Each square is $\ge 0$, so the whole sum is $\ge 0$, giving $a^2 + b^2 + c^2 \ge ab + bc + ca$. Equality holds iff $a = b = c$.
Answer:
$a^2 + b^2 + c^2 \ge ab + bc + ca$, with equality exactly when $a = b = c$.
Warm-up
Prove that $a^2 + 1 \ge 2a$ for every real number $a$.
1
Move everything to one side: the claim is equivalent to $a^2 - 2a + 1 \ge 0$.
2
Recognize the left side as a perfect square: $a^2 - 2a + 1 = (a - 1)^2$.
3
A real square is $\ge 0$, so $(a - 1)^2 \ge 0$ holds, with equality iff $a = 1$.
Answer:
$a^2 + 1 \ge 2a$, equality iff $a = 1$.
Contest level
Prove that $\dfrac{a}{b} + \dfrac{b}{a} \ge 2$ for all positive reals $a, b$.
1
Combine over a common denominator: $\dfrac{a}{b} + \dfrac{b}{a} - 2 = \dfrac{a^2 + b^2 - 2ab}{ab}$.
2
The numerator factors: $a^2 + b^2 - 2ab = (a - b)^2 \ge 0$, and the denominator $ab > 0$.
3
A nonnegative number over a positive number is $\ge 0$, so $\dfrac{a}{b} + \dfrac{b}{a} - 2 \ge 0$. Equality iff $a = b$.
Add $3$ to the left side and pair each $1$ with a fraction: $\sum \dfrac{a}{b+c} + 3 = \sum\left(\dfrac{a}{b+c} + 1\right) = \sum \dfrac{a + b + c}{b + c} = (a + b + c)\left(\dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b}\right)$.
2
Apply AM-HM (or Cauchy) to the three positive denominators: $\dfrac{1}{b+c} + \dfrac{1}{c+a} + \dfrac{1}{a+b} \ge \dfrac{9}{(b+c) + (c+a) + (a+b)} = \dfrac{9}{2(a + b + c)}$.
3
Therefore $\sum \dfrac{a}{b+c} + 3 \ge (a + b + c)\cdot \dfrac{9}{2(a+b+c)} = \dfrac{9}{2}$, so $\sum \dfrac{a}{b+c} \ge \dfrac{9}{2} - 3 = \dfrac{3}{2}$. Equality needs all denominators equal, i.e. $a = b = c$.
Generalization: the named inequalities form a toolkit — AM-GM, Cauchy-Schwarz, the Power-Mean chain $\text{QM} \ge \text{AM} \ge \text{GM} \ge \text{HM}$, rearrangement, Jensen (for convex/concave functions), and SOS (sum of squares). Each is ultimately a packaged 'square is nonnegative' statement.
Where it appears: a dedicated genre on every olympiad; on the AMC/AIME inequalities surface as 'find the minimum/maximum' optimization where the extremum is the equality case of a named inequality.
Pitfall: ALWAYS verify the equality case and that it is attainable under the constraints. Chaining two inequalities whose equality cases conflict yields a bound that is never reached, so the 'minimum' you report would be wrong — the single most common scoring error in inequality problems.
Spot the Signal
Look for problems where the key step is comparing quantities through standard bounds, transformations, and equality cases.
You can describe the hard part as comparing quantities through standard bounds, transformations, and equality cases, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the expression or equation that calls for inequalities, then rewrite the givens around it.
Name the relation that makes Inequalities legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Inequalities just because the surface looks familiar; verify the required condition first.
Applying Inequalities because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let inequalities reveal the algebraic structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is comparing quantities through standard bounds, transformations, and equality cases.