Algebraic Weapons

Why it works. Square $u = x + \tfrac1x$: the cross term is $2 \cdot x \cdot \tfrac1x = 2$, a constant, so $u^2 = x^2 + \tfrac{1}{x^2} + 2$. Cubing works the same way: $u^3 = x^3 + \tfrac{1}{x^3} + 3\left(x + \tfrac1x\right)$. Because the products $x \cdot \tfrac1x$ keep collapsing to $1$, every expression $x^n + x^{-n}$ is a polynomial in $u$. One caution: for real $x$, $|u| \ge 2$, since $x^2 - ux + 1 = 0$ needs $u^2 \ge 4$.

Spotting it. An even power minus $1$ is a difference of squares, and it keeps splitting.

Step 1. $3^8 - 1 = (3^4 - 1)(3^4 + 1) = 80 \cdot 82$.

Step 2. $80 = 2^4 \cdot 5$ and $82 = 2 \cdot 41$, so $3^8 - 1 = 2^5 \cdot 5 \cdot 41$ and the largest prime factor is $41$. Computing $6560$ first and then factoring it is the slow way to the same place.

Spotting it. Everything is symmetric in $x$ and $\tfrac1x$, so use $u = 4$ and the identities; never solve for $x = 2 \pm \sqrt3$.

Step 1. $x^3 + \tfrac{1}{x^3} = u^3 - 3u = 64 - 12 = 52$.

Step 2. $x^2 + \tfrac{1}{x^2} = u^2 - 2 = 14$, and squaring AGAIN, $x^4 + \tfrac1{x^4} = 14^2 - 2 = 194$.

Spotting it. The coefficients read the same forwards and backwards ($1, -7, 14, -7, 1$). That symmetry means dividing by $x^2$ produces $x + \tfrac1x$ and $x^2 + \tfrac1{x^2}$.

Step 1. $x = 0$ is not a root, so divide by $x^2$: $\left(x^2 + \tfrac1{x^2}\right) - 7\left(x + \tfrac1x\right) + 14 = 0$.

Step 2. With $u = x + \tfrac1x$: $(u^2 - 2) - 7u + 14 = 0$, i.e. $u^2 - 7u + 12 = 0$, so $u = 3$ or $u = 4$.

Step 3. $u = 3$: $x^2 - 3x + 1 = 0$, so $x = \tfrac{3 \pm \sqrt5}{2}$. $u = 4$: $x^2 - 4x + 1 = 0$, so $x = 2 \pm \sqrt3$. All four solutions are real: a quartic became two quadratics.

The equality case is the useful half. To minimise $x + \dfrac{9}{x}$ for $x > 0$: AM–GM gives $x + \tfrac9x \ge 2\sqrt9 = 6$, attained when $x = \tfrac9x$, i.e. $x = 3$. Any optimisation whose answer "looks symmetric" is usually AM–GM in disguise, and the equality condition tells you where the optimum sits without calculus.

With $b_i = 1$ it gives the power-mean bound $n\sum a_i^2 \ge \left(\sum a_i\right)^2$, which turns "the sum is fixed" into "the sum of squares is at least this much" — exactly the step needed for extremal problems on contests. Equality needs the vectors proportional, which again names the optimiser.

Step 1 — AM–GM on the two terms: $4x + \tfrac{25}{x} \ge 2\sqrt{4x\cdot\tfrac{25}{x}} = 2\sqrt{100} = 20$.

Step 2 — equality when $4x = \tfrac{25}{x}$, i.e. $x^2 = \tfrac{25}{4}$, $x = \tfrac52$.

Step 3 — check: $4\cdot\tfrac52 + \tfrac{25}{5/2} = 10 + 10 = 20$ ✓

Spotting it. AM–GM on the two terms gives $2\sqrt{16x}$, which still contains $x$: useless. Split $\tfrac{16}{x}$ into two equal halves so that the PRODUCT of the three terms is constant.

Step 1. $x^2 + \tfrac8x + \tfrac8x \ge 3\sqrt[3]{x^2 \cdot \tfrac8x \cdot \tfrac8x} = 3\sqrt[3]{64} = 12$.

Step 2. Equality needs $x^2 = \tfrac8x$, i.e. $x = 2$, and then $4 + 8 = 12$ ✓. The least value is $12$.

How to choose the split. The powers of $x$ in the pieces must cancel: $x^2 \cdot x^{-1} \cdot x^{-1} = x^0$.

Spotting it. A sum of fractions against a fixed sum of denominators is Cauchy–Schwarz in the form $\left(\tfrac{a^2}{x} + \tfrac{b^2}{y}\right)(x + y) \ge (a + b)^2$.

Step 1. With $a = 1$, $b = 2$: $\left(\tfrac1x + \tfrac4y\right) \cdot 1 \ge (1 + 2)^2 = 9$.

Step 2. Equality needs $\tfrac{a}{x} = \tfrac{b}{y}$, so $y = 2x$; with $x + y = 1$ that is $x = \tfrac13$, $y = \tfrac23$, and indeed $3 + 6 = 9$ ✓. The least value is $9$.

Why it works. $(\sqrt x + \sqrt y)^2 = x + y + 2\sqrt{xy}$, so a nested radical of the form $\sqrt{m + 2\sqrt n}$ denests exactly when two numbers have sum $m$ and product $n$, which is a Vieta question. Conjugates work because $(\sqrt a + \sqrt b)(\sqrt a - \sqrt b) = a - b$ has no radicals left: a difference of squares again.

Spotting it. First make the inner coefficient a $2$: $6\sqrt2 = 2\sqrt{18}$. Now look for two numbers with sum $11$ and product $18$.

Step 1. $9 + 2 = 11$ and $9 \cdot 2 = 18$.

Step 2. So $\sqrt{11 + 6\sqrt2} = \sqrt9 + \sqrt2 = 3 + \sqrt2$. Check: $(3 + \sqrt2)^2 = 9 + 6\sqrt2 + 2$ ✓

Spotting it. Each denominator is a sum of square roots of consecutive integers, and its conjugate has product $(k + 1) - k = 1$.

Step 1. $\dfrac{1}{\sqrt k + \sqrt{k + 1}} = \dfrac{\sqrt{k + 1} - \sqrt k}{(k + 1) - k} = \sqrt{k + 1} - \sqrt k$.

Step 2. The sum telescopes: $(\sqrt2 - \sqrt1) + (\sqrt3 - \sqrt2) + \cdots + (\sqrt{100} - \sqrt{99}) = \sqrt{100} - 1 = 9$.

Spotting it. The inner radical has no factor $2$ in front, and $\sqrt{21} = 2\sqrt{21/4}$ leads to fractions with no nice sum-product pair. Multiply inside and outside by $2$ instead.

Step 1. $\sqrt{5 + \sqrt{21}} = \dfrac{\sqrt{10 + 2\sqrt{21}}}{\sqrt2}$.

Step 2. $7 + 3 = 10$ and $7 \cdot 3 = 21$, so $\sqrt{10 + 2\sqrt{21}} = \sqrt7 + \sqrt3$.

Step 3. $\dfrac{\sqrt7 + \sqrt3}{\sqrt2} = \dfrac{\sqrt{14} + \sqrt6}{2}$.