Difference between revisions of "1983 IMO Problems/Problem 6"
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Determine when equality occurs. | Determine when equality occurs. | ||
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+ | ==Solution 1== | ||
+ | |||
+ | By Ravi substitution, let <math>a = y+z</math>, <math>b = z+x</math>, <math>c = x+y</math>. Then, the triangle condition becomes <math>x, y, z > 0</math>. After some manipulation, the inequality becomes: | ||
+ | |||
+ | <math>xy^3 + yz^3 + zx^3 \geq xyz(x+y+z)</math>. | ||
+ | |||
+ | By Cauchy, we have: | ||
+ | |||
+ | <math>(xy^3 + yz^3 + zx^3)(z+x+y) \geq xyz(y+z+x)^2</math> with equality if and only if <math>\frac{xy^3}{z} = \frac{yz^3}{x} =\frac{zx^3}{y}</math>. So the inequality holds with equality if and only if x = y = z. Thus the original inequality has equality if and only if the triangle is equilateral. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Without loss of generality, let <math>a \geq b \geq c > 0</math>. By Muirhead or by AM-GM, we see that <math>a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>. | ||
+ | |||
+ | If we can show that <math>a^3 b + b^3 c+ c^3 a \geq a^3 c + b^3 a + c^3 b</math>, we are done, since then <math>2(a^3 b + b^3 c+ c^3 a ) \geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>, and we can divide by <math>2</math>. | ||
+ | |||
+ | We first see that, <math>(a^2 + ac + c^2) \geq (b^2 + bc + c^2)</math>, so <math>(a-c)(b-c)(a^2 + ac + c^2) \geq (a-c)(b-c)(b^2 + bc + c^2)</math>. | ||
+ | |||
+ | Factoring, this becomes <math>(a^3 - c^3)(b-c) \geq (a-c)(b^3 - c^3)</math>. This is the same as: | ||
+ | |||
+ | <math>(a^3 - c^3)(b-c) + (b^3 - c^3)(c-a) \geq 0</math>. | ||
+ | |||
+ | Expanding and refactoring, this is equal to <math>a^3 (b-c) + b^3(c-a) + c^3 (a-b) \geq 0</math>. (This step makes more sense going backwards.) | ||
+ | |||
+ | Expanding this out, we have | ||
+ | |||
+ | <math>a^3b + b^3 c + c^3 a \geq a^3 c + b^3 a + c^3 b</math>, | ||
+ | |||
+ | which is the desired result. | ||
+ | |||
+ | == See Also == {{IMO box|year=1983|num-b=5|after=Last Problem}} |
Revision as of 23:41, 29 January 2021
Contents
Problem 6
Let , and be the lengths of the sides of a triangle. Prove that
.
Determine when equality occurs.
Solution 1
By Ravi substitution, let , , . Then, the triangle condition becomes . After some manipulation, the inequality becomes:
.
By Cauchy, we have:
with equality if and only if . So the inequality holds with equality if and only if x = y = z. Thus the original inequality has equality if and only if the triangle is equilateral.
Solution 2
Without loss of generality, let . By Muirhead or by AM-GM, we see that .
If we can show that , we are done, since then , and we can divide by .
We first see that, , so .
Factoring, this becomes . This is the same as:
.
Expanding and refactoring, this is equal to . (This step makes more sense going backwards.)
Expanding this out, we have
,
which is the desired result.
See Also
1983 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Problem |
All IMO Problems and Solutions |