2001 USAMO Problems/Problem 3

Revision as of 17:02, 7 July 2014 by 5849206328x (talk | contribs) (Solution: official solutions)

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 \[ab + bc + ca - abc = a(b + c) + bc(1-a) \ge 0.\]

Now, without loss of generality, we assume that $b$ and $c$ are either both greater than 1 or both less than one, so $(b-1)(c-1)\ge 0$. 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} \\
&= 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*} (Error compiling LaTeX. ! Extra }, or forgotten \right.)

as desired.

See also

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

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