Difference between revisions of "1983 IMO Problems/Problem 6"
(→Solution 1) |
(→Solution 1) |
||
Line 14: | Line 14: | ||
By Cauchy, we have: | 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. | + | <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== | ==Solution 2== |
Revision as of 19:37, 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 we can divide both sides of the inequality by
, and
.
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.