Difference between revisions of "1983 IMO Problems/Problem 6"
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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>. | 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 | + | 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 + ab + b^2) \geq (b^2 + bc + c^2)</math>, so <math>(a-c)(b-c)(a^2 + ab + b^2) \geq (a-c)(b-c)(b^2 + bc + c^2)</math>. | We first see that, <math>(a^2 + ab + b^2) \geq (b^2 + bc + c^2)</math>, so <math>(a-c)(b-c)(a^2 + ab + b^2) \geq (a-c)(b-c)(b^2 + bc + c^2)</math>. |
Revision as of 20:00, 27 August 2017
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.