Difference between revisions of "1964 IMO Problems/Problem 2"
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<cmath>2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)</cmath> | <cmath>2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)</cmath> | ||
− | < | + | <math>2zx^2+2zy^2+2yx^2+2yz^2+2xy^2+2xz^2+12xyz\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz</math> |
<cmath>x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz</cmath> | <cmath>x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz</cmath> | ||
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This is true by AM-GM. We can work backwards to get that the original inequality is true. | This is true by AM-GM. We can work backwards to get that the original inequality is true. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Rearrange to get | ||
+ | <cmath>a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,</cmath> | ||
+ | which is true by Schur's inequality. |
Revision as of 19:45, 16 June 2015
Problem
Suppose are the sides of a triangle. Prove that
Solution
We can use the substitution , , and to get
This is true by AM-GM. We can work backwards to get that the original inequality is true.
Solution 2
Rearrange to get which is true by Schur's inequality.