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

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==Problem 6==
 
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Let <math>a</math>, <math>b</math> and <math>c</math> be the lengths of the sides of a triangle. Prove that
 
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<math>a^2 b(a-b) + b^2 c(b-c) + c^2 (c-a) \geq 0</math>.
 
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Determine when equality occurs.
 
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==Solution 1==
 
==Solution 1==
  
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By Cauchy, we have:  
 
By Cauchy, we have:  
  
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<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.
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<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.
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==Solution 2==
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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>.
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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>.
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We first see that, <math>(a^2 + ac + c^2) \geq (b^2 + bc + c^2)</math>, so <math>(a-c)(b-c)(a^2 + ac + c^2) \geq (a-c)(b-c)(b^2 + bc + c^2)</math>.
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Factoring, this becomes <math>(a^3 - c^3)(b-c) \geq (a-c)(b^3 - c^3)</math>. This is the same as:
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<math>(a^3 - c^3)(b-c) + (b^3 - c^3)(c-a) \geq 0</math>.
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Expanding and refactoring, this is equal to <math>a^3 (b-c) + b^3(c-a) + c^3 (a-b) \geq 0</math>. (This step makes more sense going backwards.)
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Expanding this out, we have
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<math>a^3b + b^3 c + c^3 a \geq a^3 c + b^3 a + c^3 b</math>,
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which is the desired result.
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==Solution 3==
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Let <math>s</math> be the semiperimeter, <math>\frac{a+b+c}{2}</math>, of the triangle. Then, <math>a=s-\frac{-a+b+c}{2}</math>, <math>b=s-\frac{a-b+c}{2}</math>, and <math>c=s-\frac{a+b-c}{2}</math>. We let <math>x=\frac{-a+b+c}{2},</math> <math>y=\frac{a-b+c}{2}</math>, and <math>z=\frac{a+b-c}{2}.</math> (Note that <math>x,y,z</math> are all positive, since all sides must be shorter than the semiperimeter.) Then, we have <math>a=s-x</math>, <math>b=s-y</math>, and <math>c=s-z</math>. Note that <math>x+y+z=s</math>, so <cmath>a=y+z,b=x+z,c=x+y.</cmath> Plugging this into <cmath>a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq0</cmath> and doing some expanding and cancellation, we get <cmath>2x^3z+2xy^3+2yz^3-2x^2yz-2xy^2z-2xyz^2\geq0.</cmath> The fact that each term on the left hand side has at least two variables multiplied motivates us to divide the inequality by <math>2xyz</math>, which we know is positive from earlier so we can maintain the sign of the inequality. This gives <cmath>\frac{x^2}{y}-x+\frac{y^2}{z}-y-z+\frac{z^2}{x}\geq0.</cmath> We move the negative terms to the right, giving <cmath>\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z.</cmath> We rewrite this as <cmath>\sum_{cyc}\frac{x^2}{y}\geq\sum_{cyc}rx+(1-r)y.</cmath> where <math>r</math> is any real number. (This works because if we evaulate the cyclic sum, then as long as the coefficients of <math>x</math> and <math>y</math> on the right sum to 1 the right side will be <math>x+y+z</math>.
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==Solution4==
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"Solution from 111 problems in Algebra and Number Theory"

Latest revision as of 07:40, 22 May 2024

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, $(a^2 + ac + c^2) \geq (b^2 + bc + c^2)$, so $(a-c)(b-c)(a^2 + ac + c^2) \geq (a-c)(b-c)(b^2 + bc + c^2)$.

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

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

which is the desired result.

Solution 3

Let $s$ be the semiperimeter, $\frac{a+b+c}{2}$, of the triangle. Then, $a=s-\frac{-a+b+c}{2}$, $b=s-\frac{a-b+c}{2}$, and $c=s-\frac{a+b-c}{2}$. We let $x=\frac{-a+b+c}{2},$ $y=\frac{a-b+c}{2}$, and $z=\frac{a+b-c}{2}.$ (Note that $x,y,z$ are all positive, since all sides must be shorter than the semiperimeter.) Then, we have $a=s-x$, $b=s-y$, and $c=s-z$. Note that $x+y+z=s$, so \[a=y+z,b=x+z,c=x+y.\] Plugging this into \[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq0\] and doing some expanding and cancellation, we get \[2x^3z+2xy^3+2yz^3-2x^2yz-2xy^2z-2xyz^2\geq0.\] The fact that each term on the left hand side has at least two variables multiplied motivates us to divide the inequality by $2xyz$, which we know is positive from earlier so we can maintain the sign of the inequality. This gives \[\frac{x^2}{y}-x+\frac{y^2}{z}-y-z+\frac{z^2}{x}\geq0.\] We move the negative terms to the right, giving \[\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z.\] We rewrite this as \[\sum_{cyc}\frac{x^2}{y}\geq\sum_{cyc}rx+(1-r)y.\] where $r$ is any real number. (This works because if we evaulate the cyclic sum, then as long as the coefficients of $x$ and $y$ on the right sum to 1 the right side will be $x+y+z$.

Solution4

"Solution from 111 problems in Algebra and Number Theory"

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