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

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Without loss of generality, let <math>a \geq b \geq c > 0</math>. By Muirhead or by AM-GM, we see that <math>a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>.
 
Without loss of generality, let <math>a \geq b \geq c > 0</math>. By Muirhead or by AM-GM, we see that <math>a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>.
  
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If we can show that <math>a^3 b + b^3 c+ c^3 a + c^3 b \geq a^3 c +  b^3 a + c^3 b</math>, we are done, since then we can divide both sides of the inequality by <math>2</math>, and <math>2(a^3 b + b^3 c+ c^3 a + c^3 b) \geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b</math>.
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If we can show that <math>a^3 b + b^3 c+ c^3 a \geq a^3 c +  b^3 a + c^3 b</math>, we are done, since then <math>2(a^3 b + b^3 c+ c^3 a ) \geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>, and we can divide by <math>2</math>.  
  
 
We first see that, <math>(a^2 + ab + b^2) \geq (b^2 + bc + c^2)</math>, so <math>(a-c)(b-c)(a^2 + ab + b^2) \geq (a-c)(b-c)(b^2 + bc + c^2)</math>.
 
We first see that, <math>(a^2 + ab + b^2) \geq (b^2 + bc + c^2)</math>, so <math>(a-c)(b-c)(a^2 + ab + b^2) \geq (a-c)(b-c)(b^2 + bc + c^2)</math>.

Revision as of 19:00, 27 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.

Solution 2

Without loss of generality, let $a \geq b \geq c > 0$. By Muirhead or by AM-GM, we see that $a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$.

If we can show that $a^3 b + b^3 c+ c^3 a \geq a^3 c + b^3 a + c^3 b$, we are done, since then $2(a^3 b + b^3 c+ c^3 a ) \geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$, and we can divide by $2$.

We first see that, $(a^2 + ab + b^2) \geq (b^2 + bc + c^2)$, so $(a-c)(b-c)(a^2 + ab + b^2) \geq (a-c)(b-c)(b^2 + bc + c^2)$.

Factoring, this becomes $(a^3 - c^3)(b-c) \geq (a-c)(b^3 - c^3)$. This is the same as:

$(a^3 - c^3)(b-c) + (b^3 - c^3)(c-a) \geq 0$.

Expanding and refactoring, this is equal to $a^3 (b-c) + b^3(c-a) + c^3 (a-b) \geq 0$. (This step makes more sense going backwards.)

Expanding this out, we have

$a^3b + b^3 + c^3 a \geq a^3 c + b^3 a + c^3 b$,

which is the desired result.

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