Difference between revisions of "1964 IMO Problems/Problem 2"
| Line 22: | Line 22: | ||
which is true by Schur's inequality. | which is true by Schur's inequality. | ||
| − | == See Also == {{IMO box|year= | + | == See Also == {{IMO box|year=1964|num-b=1|num-a=3}} |
Revision as of 11:47, 29 January 2021
Contents
Problem
Suppose 
are the sides of a triangle. Prove that
![\[a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.\]](http://latex.artofproblemsolving.com/2/c/8/2c8afb405fe890dd89759ea3a2ed2e527038adfc.png)
Solution
We can use the substitution 
, 
, and 
to get
![\[2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)\]](http://latex.artofproblemsolving.com/6/5/2/6522426f4f40387f35379270676d7bcd6ef21add.png)
![\[x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz\]](http://latex.artofproblemsolving.com/4/4/c/44c76b06c472625ded2773717bcfa933af468183.png)
![\[\frac{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{6}\geq xyz=\sqrt[6]{x^2yx^2zy^2xy^2zz^2xz^2y}\]](http://latex.artofproblemsolving.com/d/7/c/d7cdaef5d710f396683d10490561a1d321effebb.png)
This is true by AM-GM. We can work backwards to get that the original inequality is true.
Solution 2
Rearrange to get
![\[a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,\]](http://latex.artofproblemsolving.com/0/d/c/0dcd3e5fa64e8656215b2e2662a9f90f74cd08cd.png)
which is true by Schur's inequality.
See Also
| 1964 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
