Algebra

AM ≥ GM — the one-line minimum

For positive quantities: $\frac{u+v}{2} \ge \sqrt{uv} \quad\text{with equality iff } u=v$ Any expression $ax+\frac bx$ (for $x>0$) has minimum $2\sqrt{ab}$ at $x=\sqrt{b/a}$: $x+\frac4x \ge 2\sqrt4 = 4$; $4x+\frac9x \ge 2\sqrt{36} = 12$ at $x = \frac32$. The same idea handles substitution shapes: $(x-1)^2+\frac9{(x-1)^2} \ge 2\sqrt9 = 6$, since $(x-1)^2$ is positive for $x\ne1$. Three-term version: $x^2+\frac{250}x = x^2+\frac{125}x+\frac{125}x \ge 3\sqrt[3]{125\times125} = 75$ — split the extra term so all three pieces multiply to a constant.

Completing the square — the quadratic extreme

$ax^2+bx+c = a\left(x+\frac b{2a}\right)^2 + \left(c-\frac{b^2}{4a}\right)$ $a>0$: the square is $\ge0$, so the minimum is $c-\frac{b^2}{4a}$ at $x=-\frac b{2a}$. $x^2+4x+5 = (x+2)^2+1 \ge 1$. $a<0$: maximum at the same vertex: $5+12x-3x^2 = -3(x-2)^2+17 \le 17$. Remember the two line-readings: least value for upward parabolas, greatest for downward ones — and check which one the question wants.

Fixed sum / fixed product

• Fixed sum, product maximised: equal split. $x+y = 20$ → $xy \le 100$ at $x=y=10$.
• Fixed product, sum minimised: equal factors. $xy = 36$ → $x+y \ge 12$ at $x=y=6$. Both are AM ≥ GM in disguise ($\frac{x+y}{2}\ge\sqrt{xy}$), and both are asked verbatim.

Positive definite and the discriminant

$ax^2+bx+c>0$ for ALL real $x$ needs $a>0$ AND $b^2-4ac<0$: $4x^2-kx+25>0$ always → $k^2<400$ → largest integer $k = 19$. Root-nature reads the same table: real & distinct $\iff b^2>4ac$, equal $\iff b^2=4ac$, none real $\iff b^2<4ac$. For $x^2+kx+16$ to have real distinct roots, $k^2>64$ → $|k|>8$.

Chained shapes

A hard stem often folds one tool inside another: the least value of $(x+\frac1x)^2-4(x+\frac1x)+5$ — set $t = x+\frac1x \ge 2$ first (AM ≥ GM), then $(t-2)^2+1 \ge 1$ (vertex). Two moves, each a standard one.

Common traps

• Applying AM ≥ GM to $x+\frac4x$ without checking $x>0$ (for $x<0$ the expression is negative).
• Forgetting the vertex $x$-value when the question asks WHERE the minimum occurs, not just its value.
• Using $c-\frac{b^2}{4a}$ for a downward parabola as a minimum — that shape has no minimum.
• Missing that equality in AM ≥ GM needs the two pieces EQUAL, which pins $x$.

Quick revision

• $ax+\frac bx \ge 2\sqrt{ab}$ for $x>0$; equality at $x=\sqrt{b/a}$.
• Vertex: $x=-\frac b{2a}$; extreme value $c-\frac{b^2}{4a}$ (min if $a>0$, max if $a<0$).
• Fixed sum → max product at equal split; fixed product → min sum at equal factors.
• Always positive: $a>0$ and $b^2<4ac$.
• Substitute $t = x+\frac1x \ge 2$ before completing the square.