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>
  
<cmath>2zx^2+4xyz+2zy^2+2yx^2+4xyz+2yz^2+2xy^2+4xyz+2xz^2\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz</cmath>
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<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.
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==Solution 2==
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Rearrange to get
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<cmath>a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,</cmath>
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which is true by Schur's inequality.
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== See Also == {{IMO box|year=1964|num-b=1|num-a=3}}

Revision as of 12:47, 29 January 2021

Problem

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

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

Solution

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.

Solution 2

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.

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