Deep-Dive Week 1: Your 2 Weakest Topics

Day one of a rebuild: write the topic's spine from memory — definitions, then theorems in dependency order, then the proof IDEA of each in one line ('MVT: tilt Rolle by subtracting the secant line'). Where you cannot produce the idea, that theorem was memorized, not owned, and it marks exactly where your problem-solving fails under pressure: you cannot adapt a proof whose mechanism you never understood.

Days two and three: re-solve the topic's canonical problems — the ones from Weeks 1-9 you got RIGHT — with notes closed. Right-with-notes and right-cold are different states of knowledge, and the second is the only one that will exist in the exam room. Expect an uncomfortable fraction to have evaporated; that fraction is the honest size of your weakness.

Days four onward: fresh problems, but with a modified debrief. After each attempt, trace the failure (or success) back to a SPECIFIC line of the spine you wrote on day one. 'Stalled because I forgot compactness gives uniform continuity' is a spine repair; over a week, the annotated spine becomes the densest study artifact you own.

Pitfall. Rebuilding by rereading. Passive review feels like progress and measures nothing; every step of the protocol above happens with the book CLOSED, notes opened only to check.

Rearrangement inequality: for sorted reals $a_1 \le \cdots \le a_n$ and $b_1 \le \cdots \le b_n$, the sum $\sum a_i b_{\sigma(i)}$ is maximized by the identity pairing (sorted-with-sorted) and minimized by the reversal. No positivity needed — this is the most hypothesis-free inequality you own, and it is the reason so many symmetric inequalities let you ASSUME an ordering $a \ge b \ge c$ for free.

Chebyshev's sum inequality is its averaged corollary: same-sorted sequences satisfy $\frac{1}{n}\sum a_i b_i \ge \left(\frac{1}{n}\sum a_i\right)\left(\frac{1}{n}\sum b_i\right)$, reversed for oppositely-sorted. Read it as a correlation statement — similarly ordered sequences are positively correlated — and it becomes easy to remember and easy to spot: any 'sum of products vs product of sums' comparison where the sequences move together.

The tactical habit: when a cyclic or symmetric inequality resists AM-GM, ask 'are both sides sums of products of things I can sort?' Rearrangement handles individual pairings, Chebyshev handles the averages, and together they resolve a class of problems that pure convexity tools fumble.

Pitfall. Assuming an ordering in a CYCLIC (not symmetric) inequality. Cyclic expressions only allow rotating the variables, so you may fix which variable is largest but not fully sort — check which symmetry the problem actually has before writing 'WLOG $a \ge b \ge c$'.

Schur's inequality: for $x, y, z \ge 0$ and $t > 0$, $x^t(x-y)(x-z) + y^t(y-x)(y-z) + z^t(z-x)(z-y) \ge 0$. At $t = 1$ it expands to $x^3 + y^3 + z^3 + xyz \cdot 3 \ge xy(x+y) + yz(y+z) + zx(z+x)$ — the inequality that rescues you precisely when AM-GM dies because the needed term ($xyz$) sits on the WRONG side. Proof idea worth owning: sort $x \ge y \ge z$ and group the first two terms; each group is visibly nonnegative.

SOS (sum of squares): rewrite the difference of sides as $\sum S_c (a-b)^2$ with coefficient functions $S_c$. If all $S_c \ge 0$ you are done; if not, standard lemmas (e.g. two adjacent coefficients nonnegative with the right dominance) still close it. SOS is the inequality analogue of 'show the discriminant is negative' — mechanical, powerful, and gradeable, because every step is an identity plus a sign check.

Homogenization and normalization are inverse gears. A constraint like $a + b + c = 3$ lets you scale it INTO the inequality (replace constants by powers of $\frac{a+b+c}{3}$) until every term has equal degree — then the constraint disappears and degree-based tools apply. Normalization runs the other way: a homogeneous inequality lets you IMPOSE $a + b + c = 3$ or $abc = 1$, whichever collapses the mess. Smoothing completes the set: show that moving two variables toward their mean (holding the constraint) pushes the expression the right way; iterate, and the extremum is forced to the all-equal point — or, when smoothing fails at a boundary, to an edge where a variable is 0, which is a smaller problem you solve directly.

Pitfall. Expecting one named inequality to finish alone. Contest inequalities are usually two-tool problems — Schur plus AM-GM here — and if a chain stalls, the smoothing reduction to 'two variables equal' is the universal fallback for symmetric three-variable problems.