Algebra & Inequalities

For positive numbers: $\dfrac{a_1+\cdots+a_n}{n}\ge\sqrt[n]{a_1\cdots a_n}$ (AM–GM), with equality if and only if all are equal.

Power mean chain: $\text{(root mean square)}\ge\text{(arithmetic)}\ge\text{(geometric)}\ge\text{(harmonic)}$. Particularly useful is $a^2+b^2+c^2\ge\dfrac{(a+b+c)^2}{3}$ (QM–AM).

Technique: when the product of a set of numbers is constant (especially 1), apply AM–GM directly — a product of 1 makes the geometric mean $=1$, and the lower bound appears at once (e.g. $\dfrac ab+\dfrac bc+\dfrac ca\ge3$). Problems with a product constraint $abc=1$ often use the two-step squeeze of "QM–AM to bound the sum of squares + AM–GM to bound $\sum a\ge3$".

Sum of squares being non-negative is the most elementary source of inequalities:

To prove a polynomial is always positive (when there is no unified completing of the square): split according to the range of the variable, and on each segment reorganise the expression into a sum of obviously non-negative blocks (the same expression uses different groupings on different intervals). For inequalities with a large span, use two-level SOS: first reduce the quartic to "products of squares", then reduce to the target (e.g. $a^4+b^4+c^4\ge abc(a+b+c)$).

$\left(\sum a_ib_i\right)^2\le\left(\sum a_i^2\right)\left(\sum b_i^2\right)$, with equality when the two sequences are proportional. One-line proof: $Q(t)=\sum(a_it-b_i)^2\ge0$, discriminant $\le0$.

Engel / Titu form (the fraction workhorse): $\displaystyle\sum\frac{x_i^2}{y_i}\ge\frac{(\sum x_i)^2}{\sum y_i}$. Combined with homogenising by "adding 1 to each term", this proves Nesbitt $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac32$.

Let the power sums be $p_k=\sum x_i^k$ and the elementary symmetric polynomials be $e_1,e_2,e_3$. Newton's identities express $p_k$ recursively as a polynomial in $e$: $p_1=e_1$, $p_2=e_1p_1-2e_2$, $p_k=e_1p_{k-1}-e_2p_{k-2}+e_3p_{k-3}$.

With constraints there is massive collapse: $a+b+c=0$ ($e_1=0$) gives $p_2=-2e_2,\ p_3=3e_3,\ p_5=-5e_2e_3$, hence $\dfrac{p_5}{5}=\dfrac{p_2}{2}\cdot\dfrac{p_3}{3}$.

The single-variable symmetric expression $x^k+\dfrac1{x^k}$ satisfies $p_k=s\,p_{k-1}-p_{k-2}$ ($s=x+\tfrac1x$).

Crafting a difference of squares (Sophie Germain's idea): $n^4+n^2+1=(n^2+n+1)(n^2-n+1)$, $n^4+4=(n^2-2n+2)(n^2+2n+2)$, used to prove that something that looks prime is in fact composite.

Rational root theorem: a rational root $\tfrac pq$ of a polynomial with integer coefficients satisfies $p\mid$ the constant term, $q\mid$ the leading coefficient — used to prove that an algebraic number is irrational (e.g. a real root of $x^3=x+1$).

Sequences: when a recurrence of the form $a_{n+1}=a_n+f(a_n)$ is hard to analyse directly, instead look at $a_n^2$ (or another monotone quantity) to linearise the recurrence (e.g. $a_{n+1}^2=a_n^2+2+\tfrac1{a_n^2}$ gives $a_n^2\ge2n-1$). Monotone bounded convergence: first prove bounded + monotone ⇒ the limit exists, then take the limit of the recurrence to solve for $L$.

Common techniques: substitute special values ($x=y=0,1$ determines $f(0),f(1)$), exploit symmetry/substitution, upgrade preservation of sign to monotonicity.

Classic result: the only $f:\mathbb R\to\mathbb R$ that is both additive and multiplicative are $f\equiv0$ and $f=\mathrm{id}$ — the breakthrough is that $f(x)=f(\sqrt x)^2\ge0$ ($x\ge0$) yields 'preservation of sign ⇒ monotonicity', and then the density of rational points pins it down to the identity.