Olympiad Inequalities II: The Full Ladder

The constraint tells you how to group: a product constraint suggests AM–GM across terms whose product is a power of it; a sum constraint suggests QM–AM or the identity $(a + b + c)^2 = \sum a^2 + 2\sum ab$. Homogenize first when degrees differ — replace a constant by the appropriate power of $(a + b + c)$ or $(abc)^{1/3}$ — so that the inequality becomes scale-free and the standard tools apply.

Engel form is Cauchy–Schwarz with $u_i = \tfrac{a_i}{\sqrt{b_i}}$, $v_i = \sqrt{b_i}$; it is the fastest route to bounds on sums of fractions. Equality in Engel: $\tfrac{a_i}{b_i}$ all equal — so the minimizer of $\tfrac1x + \tfrac4y + \tfrac9z$ has $x : y : z = 1 : 2 : 3$. Write the equality case down before computing; it often IS the answer.

Schur handles the inequalities AM–GM cannot (it is the one with equality at $a = b = c$ AND at $(a, a, 0)$). SOS is a systematic finishing move for symmetric three-variable inequalities: expand, group into squares. Smoothing (or "mixing variables") proves that the extremum occurs when variables are equal or when one hits the boundary — reducing a three-variable problem to one variable. These three are the difference between AIME-level and USAMO-level inequality work.

Homogenize: $1 + \tfrac1a = \tfrac{a + b + c + a}{a}$... simpler: $1 + \tfrac1a = 1 + \tfrac{a + b + c}{a} = \tfrac{2a + b + c}{a}$. By AM–GM $2a + b + c = a + a + b + c \ge 4\sqrt[4]{a^2bc}$; multiply the three: numerator $\ge 64\sqrt[4]{a^4b^4c^4} = 64abc$, denominator $abc$. Equality at $a = b = c = \tfrac13$ ✓.

By symmetry assume $a \ge b \ge c$. Group the first two terms: $(a - b)\left[a(a - c) - b(b - c)\right] \ge 0$ because $a - b \ge 0$ and $a(a - c) \ge b(b - c)$ (both factors on the left dominate). The third term $c(c - a)(c - b) = c(a - c)(b - c) \ge 0$ is a product of nonnegatives. $\blacksquare$ Equality iff $a = b = c$ or two are equal and the third is $0$ — the two-equality-case signature that makes Schur indispensable when AM–GM's single equality case is too restrictive.

If $\sigma$ is not the identity, some $i < j$ has $\sigma(i) > \sigma(j)$; swapping the two assignments changes the sum by $(a_j - a_i)(b_{\sigma(i)} - b_{\sigma(j)}) \ge 0$. Repeated swaps reach the identity without ever decreasing the sum. $\blacksquare$ Significance: Chebyshev's inequality (similarly ordered sequences have $\tfrac1n\sum a_ib_i \ge \bar a\bar b$) is rearrangement averaged over all cyclic shifts, and both are the tool when AM–GM's symmetry is broken by an ordering assumption.

Step 1 — $2(a^2 + b^2 + c^2) - 2(ab + bc + ca) = (a - b)^2 + (b - c)^2 + (c - a)^2 \ge 0$: a sum of squares.

Step 2 — equality iff every square vanishes iff $a = b = c$.

Step 3 — divide by the positive $ab + bc + ca$.

Step 4 — the same SOS identity gives $(a + b + c)^2 \ge 3(ab + bc + ca)$ and $3(a^2 + b^2 + c^2) \ge (a + b + c)^2$ — three inequalities from one line of algebra, all sharing the equality case $a = b = c$.

Step 1 — test the symmetric point $a = b = c = 1$: value $3 - 1 = 2$. Test a boundary point $(1.5, 1.5, 0)$: $2.25$. So the maximum is NOT at the symmetric point — smoothing must be done carefully.

Step 2 — fix $c$ and set $a + b = 3 - c = s$; the expression is $ab(1 - c) + cs$. For $c < 1$ it increases with $ab$, maximized at $a = b$; for $c > 1$ it is maximized at $ab = 0$.

Step 3 — case $a = b = \tfrac s2$: $f(c) = \tfrac{(3-c)^2}{4}(1 - c) + c(3 - c)$ on $[0, 1]$; $f(0) = 2.25$, $f(1) = 2$, and $f' < 0$ on the interval — max $2.25$ at $c = 0$. Case $ab = 0$ with $c \ge 1$: value $c(3 - c) \le 2.25$.

Step 4 — maximum $\tfrac94$ at $(\tfrac32, \tfrac32, 0)$ and permutations. Boundary cases are where symmetric-point intuition fails; smoothing finds them systematically.