Solving Problems by Constructing a Function

In one sentence

The problem does not mention a function, so you build one: pick one letter as the variable and read the expression as the value of that function at some point. Then “prove an inequality” or “solve an equation” turns into “look at the ends of a segment” or “check whether a function is monotonic”.

The hard part is never the properties of functions (you met them all in earlier chapters). It is seeing which function to build.

Predict first

$a, b, c$ all take values in $[0, 1]$ . What is the largest possible value of $a + b + c - ab - b c - c a$ ?

\frac{3}{4}

1

\frac{3}{2}

3

Choose an answer first; then you can see the result.

Stop 1 · Inequalities: for a linear function, check the ends

Prove: if 0 ≤ a, b, c ≤ 1, then a + b + c − ab − bc − ca ≤ 1

Step 1 / 5

Take a as the variable
Freeze $b$ and $c$ and let only $a$ move. The expression has degree $1$ in $a$ : it is a linear function of $a$ , and its graph is a segment.

$g (a) = (1 - b - c) a + (b + c - b c)$
Left end a = 0
Right end a = 1
The segment lies between its ends
Equality

Stop 2 · Extreme values: build a quadratic equation

Example. Real numbers $x, y, z$ satisfy $x + y + z = 5$ and $x y + y z + z x = 3$ . Find the maximum of $z$ .

Three letters and two conditions: you cannot solve for $z$ directly. But freeze $z$ , and you know both the sum and the product of $x$ and $y$ . So they are the two roots of a quadratic equation, and “it has real roots” is an inequality in $z$ .

Build the equation whose roots are x and y

Step 1 / 4

Sum and product
$x + y = 5 - z$ , and $x y = 3 - z (x + y) = 3 - z (5 - z) = z^{2} - 5 z + 3$ .

$x + y = 5 - z, x y = z^{2} - 5 z + 3$
Build the equation
Real roots ⇔ Δ ≥ 0
Equality

The same idea proves the Cauchy inequality: $f (t) = \sum (a_{i} t - b_{i})^{2} \geq 0$ for every $t$ , so the discriminant of this quadratic must be $\leq 0$ . Written out, that is $(\sum a_{i} b_{i})^{2} \leq \sum a_{i}^{2} \sum b_{i}^{2}$ .

Stop 3 · Equations: an increasing odd function

One problem, several solutions

Solve $(2 x + 1)^{3} + (2 x + 1) + x^{3} + x = 0$ .

Expand it

Expanding: $9 x^{3} + 12 x^{2} + 9 x + 2 = 0$ .

Try $x = - \frac{1}{3}$ : $- \frac{1}{3} + \frac{4}{3} - 3 + 2 = 0$ ✓, so $(3 x + 1) (3 x^{2} + 3 x + 2) = 0$ .

The discriminant of $3 x^{2} + 3 x + 2$ is $9 - 24 < 0$ , so $x = - \frac{1}{3}$ is the only solution.

CostAlways works, but you have to factor a cubic and hope to find a rational root.

Build f(t) = t³ + t

The equation says $f (2 x + 1) + f (x) = 0$ . $f$ is odd, so $f (2 x + 1) = - f (x) = f (- x)$ .

$f$ is strictly increasing on $R$ (a sum of two increasing functions), so $2 x + 1 = - x$ and $x = - \frac{1}{3}$ .

CostDone in two lines; the key is seeing that the two groups “look the same”.

Notice that the second solution uses two properties: oddness turns “ $+$ ” into “ $=$ ”, and monotonicity turns “equal values” into “equal inputs”. Drop either one and the argument fails (see this stop’s exhibits in the Mistake Museum).

Stop 4 · Polynomials: build a polynomial with many roots

You know several values of a polynomial and want another one: turn what you know into roots.

Example (a simplified version of a 1975 USA Mathematical Olympiad problem). A cubic polynomial $f$ satisfies $f (k) = \frac{k}{k + 1}$ for $k = 0, 1, 2, 3$ . Let $g (x) = (x + 1) f (x) - x$ . It is a polynomial of degree four with $g (0) = g (1) = g (2) = g (3) = 0$ , so $g (x) = c x (x - 1) (x - 2) (x - 3)$ . The constant $c$ is fixed by $x = - 1$ : $g (- 1) = 1 = c \cdot 24$ . That is enough to compute $f$ at any point; it is left as Exercise 4.

Practice

Exercise 1Basic

$a, b \in [0, 1]$ . Find the maximum of $2 a + 2 b - ab$ .

Exercise 2Advanced

Solve $x^{5} + x + (x - 2)^{5} + (x - 2) = 0$ .

Exercise 3Advanced

$a, b, c \in [- 1, 1]$ . Find the maximum of $a + b + c - ab c$ .

This plots

ab c + 2 - (a + b + c)

with bound

0

: it is never less than

0

, which is exactly

a + b + c - ab c \leq 2

Exercise 4Challenge

A cubic polynomial $f$ satisfies $f (k) = \frac{k}{k + 1}$ ( $k = 0, 1, 2, 3$ ). Find $f (4)$ .

In one sentence

Stop 1 · Inequalities: for a linear function, check the ends

Prove: if 0 ≤ a, b, c ≤ 1, then a + b + c − ab − bc − ca ≤ 1

Stop 2 · Extreme values: build a quadratic equation

Build the equation whose roots are x and y

Stop 3 · Equations: an increasing odd function

Expand it

Build f(t) = t³ + t

Stop 4 · Polynomials: build a polynomial with many roots

Practice

Mistake Museum · Exhibits from this stop

Checking only the endpoints for a quadratic as well

Concluding $u=v$ from $f(u)=f(v)$ when $f$ is not monotonic