Difference between revisions of "1964 IMO Problems/Problem 2"

Revision as of 18:45, 16 June 2015

Suppose $a, b, c$ are the sides of a triangle. Prove that

$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.$

We can use the substitution $a=x+y$ , $b=x+z$ , and $c=y+z$ to get

$2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)$

$2zx^2+2zy^2+2yx^2+2yz^2+2xy^2+2xz^2+12xyz\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz$

$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz$

$\frac{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{6}\geq xyz=\sqrt[6]{x^2yx^2zy^2xy^2zz^2xz^2y}$

This is true by AM-GM. We can work backwards to get that the original inequality is true.

Rearrange to get $a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,$ which is true by Schur's inequality.

@@ Line 16: / Line 16: @@
 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.