Difference between revisions of "1983 IMO Problems/Problem 6"

m (Solution 1)
m (Solution 1)
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By Cauchy, we have:  
 
By Cauchy, we have:  
  
<math>(xy^3 + yz^3 + zx^3)(z+x+y) \geq xyz(y+z+x)^2</math> with equality if and only if <math>\frac{xy^3}{z} = \frac{yz^3}{x} = \frac{zx^3}{y}</math>. So the inequality holds with equality if and only if x = y = z. Thus the original inequality has equality if and only if the triangle is equilateral.
+
<math>(xy^3 + yz^3 + zx^3)(z+x+y) \geq xyz(y+z+x)^2</math> with equality if and only if <math>\frac{xy^3}{z} = \frac{yz^3}{x} = \frac{zx^3}{y}</math>. So the inequality holds with equality if and only if <math>x = y = z</math>. Thus the original inequality has equality if and only if the triangle is equilateral.

Revision as of 17:50, 22 August 2017

Problem 6

Let $a$, $b$ and $c$ be the lengths of the sides of a triangle. Prove that

$a^2 b(a-b) + b^2 c(b-c) + c^2 (c-a) \geq 0$.

Determine when equality occurs.

Solution 1

By Ravi substitution, let $a = y+z$, $b = z+x$, $c = x+y$. Then, the triangle condition becomes $x, y, z > 0$. After some manipulation, the inequality becomes:

$xy^3 + yz^3 + zx^3 \geq xyz(x+y+z)$.

By Cauchy, we have:

$(xy^3 + yz^3 + zx^3)(z+x+y) \geq xyz(y+z+x)^2$ with equality if and only if $\frac{xy^3}{z} = \frac{yz^3}{x} = \frac{zx^3}{y}$. So the inequality holds with equality if and only if $x = y = z$. Thus the original inequality has equality if and only if the triangle is equilateral.

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