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Difference between revisions of "1964 IMO Problems/Problem 2"
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~mathboy100
 
~mathboy100
  
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== Solution 2 ==
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== Solution 2 (Ravi Substitution) ==
 
We can use the substitution <math>a=x+y</math>, <math>b=x+z</math>, and <math>c=y+z</math> to get
 
We can use the substitution <math>a=x+y</math>, <math>b=x+z</math>, and <math>c=y+z</math> 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>
 
<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.
 
which is true by Schur's inequality.
 +
 +
==Video Solution==
 +
https://youtu.be/6gDLBT1aGQM?si=ZR78mdotq4wfS3SA&t=637 [little fermat]
  
 
== See Also == {{IMO box|year=1964|num-b=1|num-a=3}}
 
== See Also == {{IMO box|year=1964|num-b=1|num-a=3}}

Latest revision as of 18:03, 15 March 2025

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

Let $b+c-a = x$, $c+a-b = y$, and $a+b-c = z$. Then, $a = \frac{y+z}{2}$, $b = \frac{x+z}{2}$, and $c = \frac{x+y}{2}$. By AM-GM, \[\frac{x+y}{2} \geq \sqrt{xy},\] \[\frac{y+z}{2} \geq \sqrt{yz},\] \[\textrm{and }\frac{x+z}{2} \geq \sqrt{xz}.\]

Multiplying these equations, we have \[\frac{x+y}{2} \cdot \frac{y+z}{2} \cdot \frac{x+z}{2} \geq xyz\] \[\therefore abc \geq (a+b-c)(b+c-a)(c+a-b).\] We can now simplify: \[(a+b-c)(b+c-a)(c+a-b) \leq abc\] \[(-a^2 + b^2 - c^2 + 2ac)(c+a-b) \leq abc\] \[a(-a^2 + b^2 - c^2 + 2ac) + c(-a^2 + b^2 - c^2 + 2ac) - b(-a^2 + b^2 - c^2 + 2ac) \leq abc\] \[-a^3 + ab^2 - ac^2 + 2a^2c - a^2c + b^2c - c^3 + 2ac^2 + a^2b - b^3 + bc^2 - 2abc \leq abc\] \[a^2b + a^2c - a^3 + b^2c + ab^2 - b^3 + ac^2 + bc^2 - c^3 - 2abc \leq abc\] \[a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}\textrm{. }\square\] ~mathboy100

Solution 2 (Ravi Substitution)

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 3

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.

Video Solution

https://youtu.be/6gDLBT1aGQM?si=ZR78mdotq4wfS3SA&t=637 [little fermat]

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
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