2001 USAMO Problems/Problem 3

Problem

Let $a, b, c \geq 0$ and satisfy

$a^2 + b^2 + c^2 + abc = 4.$

Show that

$0 \le ab + bc + ca - abc \leq 2.$

Solution

Solution 1

First we prove the lower bound.

Note that we cannot have $a, b, c$ all greater than 1. Therefore, suppose $a \le 1$ . Then $ab + bc + ca - abc = a(b + c) + bc(1-a) \ge 0.$ Note that, by the Pigeonhole Principle, at least two of $a,b,c$ are either both greater than or less than $1$ . Without loss of generality, let them be $b$ and $c$ . Therefore, $(b-1)(c-1)\ge 0$ . From the given equation, we can express $a$ in terms of $b$ and $c$ as $a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2}$ Thus, $ab + bc + ca - abc = -a (b-1)(c-1)+a+bc \le a+bc = \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2}$

From the Cauchy-Schwarz Inequality, $\frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2} \le \frac{\sqrt{(4-b^2+b^2)(4-c^2+c^2)} }{2} = 2.$

This completes the proof.

Solution 2

The proof for the lower bound is the same as in the first solution.

Now we prove the upper bound. Let us note that at least two of the three numbers $a$ , $b$ , and $c$ are both greater than or equal to 1 or less than or equal to 1. Without loss of generality, we assume that the numbers with this property are $b$ and $c$ . Then we have $(1 - b)(1 - c)\geq 0.$ The given equality $a^2 + b^2 + c^2 + abc = 4$ and the inequality $b^2 + c^2\geq 2bc$ imply that $a^2 + 2bc + abc\leq 4,$ or $bc(2 + a)\leq 4 - a^2.$ Dividing both sides of the last inequality by $2 + a$ yields $bc\leq 2 - a.$ Thus, $ab + bc + ca - abc\leq ab + 2 - a + ac(1 - b) = 2 - a(1 + bc - b - c) = 2 - a(1 - b)(1 - c)\leq 2,$ as desired.

The last equality holds if and only if $b = c$ and $a(1 - b)(1 - c) = 0$ . Hence equality for the upper bound holds if and only if $(a,b,c)$ is one of the triples $(1,1,1)$ , $(0,\sqrt{2},\sqrt{2})$ , $(\sqrt{2},0,\sqrt{2})$ , and $(\sqrt{2},\sqrt{2},0)$ . Equality for the lower bound holds if and only if $(a,b,c)$ is one of the triples $(2,0,0)$ , $(0,2,0)$ and $(0,0,2)$ .

Solution 3

The proof for the lower bound is the same as in the first solution.

Now we prove the upper bound. It is clear that $a,b,c\leq 2$ . If $abc = 0$ , then the result is trivial. Suppose that $a,b,c > 0$ . Solving for $a$ yields $a = \frac{-bc + \sqrt{b^2c^2 - 4(b^2 + c^2 - 4)}}{2} = \frac{-bc + \sqrt{(4 - b^2)(4 - c^2)}}{2}.$ This asks for the trigonometric substitution $b = 2\sin u$ and $c = 2\sin v$ , where $0^\circ < u,v < 90^\circ$ . Then $a = 2(-\sin u\sin v + \cos u\cos v) = 2\cos (u + v),$ and if we set $u = B/2$ and $v = C/2$ , then $a = 2\sin (A/2)$ , $b = 2\sin (B/2)$ , and $c = \sin (C/2)$ , where $A$ , $B$ , and $C$ are the angles of a triangle. We have $\begin{align*} ab &= 4\sin\frac{A}{2}\sin\frac{B}{2} \\ &= 2\sqrt{\sin A\tan\frac{A}{2}\sin B\tan\frac{B}{2}} = 2\sqrt{\sin A\tan\frac{B}{2}\sin B\tan\frac{A}{2}} \\ &\leq \sin A\tan\frac{B}{2} + \sin B\tan\frac{A}{2} \\ &= \sin A\cot\frac{A + C}{2} + \sin B\cot\frac{B + C}{2}, \end{align*}$ where the inequality step follows from AM-GM. Likewise, $\begin{align*} bc &\leq \sin B\cot\frac{B + A}{2} + \sin C\cot\frac{C + A}{2}, \\ ca &\leq \sin A\cot\frac{A + B}{2} + \sin C\cot\frac{C + B}{2}. \end{align*}$ Therefore $\begin{align*} ab + bc + ca &\leq (\sin A + \sin B)\cot\frac{A + B}{2} + (\sin B + \sin C)\cot\frac{B + C}{2} + (\sin C + \sin A)\cot\frac{C + A}{2} \\ &= 2\left(\cos\frac{A - B}{2}\cos\frac{A + B}{2} + \cos\frac{B - C}{2}\cos\frac{B + C}{2} + \cos\frac{C - A}{2}\cos\frac{C + A}{2} \right)\\ &= 2(\cos A + \cos B + \cos C) \\ &= 6 - 4\left(\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2}\right) \\ &= 6 - (a^2 + b^2 + c^2) = 2 + abc, \end{align*}$ as desired.

2001 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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