Difference between revisions of "2001 USAMO Problems/Problem 3"
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<center><math>a^2 + b^2 + c^2 + abc = 4.</math></center> | <center><math>a^2 + b^2 + c^2 + abc = 4.</math></center> | ||
Show that | Show that | ||
− | <center><math>ab + bc + ca - abc \leq 2.</math></center> | + | <center><math>0 \le ab + bc + ca - abc \leq 2.</math></center> |
== Solution == | == Solution == | ||
− | + | === Solution 1 === | |
+ | First we prove the lower bound. | ||
− | Without | + | Note that we cannot have <math>a, b, c</math> all greater than 1. |
− | < | + | Therefore, suppose <math>a \le 1</math>. |
+ | Then | ||
+ | <cmath>ab + bc + ca - abc = a(b + c) + bc(1-a) \ge 0.</cmath> | ||
+ | Note that, by the [[Pigeonhole Principle]], at least two of <math>a,b,c</math> are either both greater than or less than <math>1</math>. [[Without loss of generality]], let them be <math>b</math> and <math>c</math>. Therefore, <math>(b-1)(c-1)\ge 0</math>. From the given equation, we can express <math>a</math> in terms of <math>b</math> and <math>c</math> as | ||
+ | <cmath>a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2}</cmath> | ||
Thus, | Thus, | ||
− | < | + | <cmath>ab + bc + ca - abc = -a (b-1)(c-1)+a+bc \le a+bc = \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2}</cmath> |
+ | |||
+ | From the [[Cauchy-Schwarz Inequality]], | ||
+ | <cmath>\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.</cmath> | ||
+ | |||
+ | 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 <math>a</math>, <math>b</math>, and <math>c</math> 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 <math>b</math> and <math>c</math>. Then we have | ||
+ | <cmath>(1 - b)(1 - c)\geq 0.</cmath> | ||
+ | The given equality <math>a^2 + b^2 + c^2 + abc = 4</math> and the inequality <math>b^2 + c^2\geq 2bc</math> imply that | ||
+ | <cmath>a^2 + 2bc + abc\leq 4,</cmath> | ||
+ | or | ||
+ | <cmath>bc(2 + a)\leq 4 - a^2.</cmath> | ||
+ | Dividing both sides of the last inequality by <math>2 + a</math> yields | ||
+ | <cmath>bc\leq 2 - a.</cmath> | ||
+ | Thus, | ||
+ | <cmath>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,</cmath> | ||
+ | as desired. | ||
+ | |||
+ | The last equality holds if and only if <math>b = c</math> and <math>a(1 - b)(1 - c) = 0</math>. Hence equality for the upper bound holds if and only if <math>(a,b,c)</math> is one of the triples <math>(1,1,1)</math>, <math>(0,\sqrt{2},\sqrt{2})</math>, <math>(\sqrt{2},0,\sqrt{2})</math>, and <math>(\sqrt{2},\sqrt{2},0)</math>. Equality for the lower bound holds if and only if <math>(a,b,c)</math> is one of the triples <math>(2,0,0)</math>, <math>(0,2,0)</math> and <math>(0,0,2)</math>. | ||
+ | |||
+ | === 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 <math>a,b,c\leq 2</math>. If <math>abc = 0</math>, then the result is trivial. Suppose that <math>a,b,c > 0</math>. Solving for <math>a</math> yields | ||
+ | <cmath>a = \frac{-bc + \sqrt{b^2c^2 - 4(b^2 + c^2 - 4)}}{2} = \frac{-bc + \sqrt{(4 - b^2)(4 - c^2)}}{2}.</cmath> | ||
+ | This asks for the trigonometric substitution <math>b = 2\sin u</math> and <math>c = 2\sin v</math>, where <math>0^\circ < u,v < 90^\circ</math>. Then | ||
+ | <cmath>a = 2(-\sin u\sin v + \cos u\cos v) = 2\cos (u + v),</cmath> | ||
+ | and if we set <math>u = B/2</math> and <math>v = C/2</math>, then <math>a = 2\sin (A/2)</math>, <math>b = 2\sin (B/2)</math>, and <math>c = \sin (C/2)</math>, where <math>A</math>, <math>B</math>, and <math>C</math> are the angles of a triangle. We have | ||
+ | <cmath>\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*}</cmath> | ||
+ | where the inequality step follows from AM-GM. Likewise, | ||
+ | <cmath>\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*}</cmath> | ||
+ | Therefore | ||
+ | <cmath>\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*}</cmath> | ||
+ | as desired. | ||
== See also == | == See also == | ||
{{USAMO newbox|year=2001|num-b=2|num-a=4}} | {{USAMO newbox|year=2001|num-b=2|num-a=4}} | ||
+ | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:45, 9 April 2023
Problem
Let and satisfy
![$a^2 + b^2 + c^2 + abc = 4.$](http://latex.artofproblemsolving.com/1/6/e/16e70ab813b2e9287a1015d7b890d16f94a7073e.png)
Show that
![$0 \le ab + bc + ca - abc \leq 2.$](http://latex.artofproblemsolving.com/b/f/7/bf76c4b7c8d7148354b55dc865be48fdeb231c1a.png)
Solution
Solution 1
First we prove the lower bound.
Note that we cannot have all greater than 1.
Therefore, suppose
.
Then
Note that, by the Pigeonhole Principle, at least two of
are either both greater than or less than
. Without loss of generality, let them be
and
. Therefore,
. From the given equation, we can express
in terms of
and
as
Thus,
From the Cauchy-Schwarz Inequality,
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 ,
, and
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
and
. Then we have
The given equality
and the inequality
imply that
or
Dividing both sides of the last inequality by
yields
Thus,
as desired.
The last equality holds if and only if and
. Hence equality for the upper bound holds if and only if
is one of the triples
,
,
, and
. Equality for the lower bound holds if and only if
is one of the triples
,
and
.
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 . If
, then the result is trivial. Suppose that
. Solving for
yields
This asks for the trigonometric substitution
and
, where
. Then
and if we set
and
, then
,
, and
, where
,
, and
are the angles of a triangle. We have
where the inequality step follows from AM-GM. Likewise,
Therefore
as desired.
See also
2001 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.