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

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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>.
 
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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Thus, we need to show that there exists a real number <math>r</math> such that <math>\frac{x^2}{y}\geq rx+(1-r)y</math> for all positive <math>x,y</math>. We claim that <math>r=2</math> works. This becomes <math>\frac{x^2}{y}\geq2x-y</math>, and since <math>y</math> is positive we can multiply by <math>y</math> to yield <math>x^2\geq2xy-y^2</math>, or <math>(x-y)^2\geq0</math>, which is obviously true by the Trivial Inequality. Thus, we are done with part (a).
 
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(To clarify how this works, we have <math>\frac{x^2}{y}\geq 2x-y</math>, <math>\frac{y^2}{z}\geq 2y-z</math>, and <math>\frac{z^2}{x}\geq 2z-x</math>, we add these inequalities to get <math>\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z.</math>)
 
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Equality occurs if and only if <math>\frac{x^2}{y}=2x-y</math>, <math>\frac{y^2}{z}=2y-z</math>, and <math>\frac{z^2}{x}=2z-x</math> at the same time, since if they are not all equal then the left side will be greater than the right side. From there, it is easy to see that the equality case is <math>x=y=z</math>, which is <math>a=b=c</math>.
 
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~john0512
 
 
==Solution4==
 
==Solution4==
 
"Solution from 111 problems in Algebra and Number Theory"
 
"Solution from 111 problems in Algebra and Number Theory"

Revision as of 06:39, 22 May 2024

Problem 6

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

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

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, $(a^2 + ac + c^2) \geq (b^2 + bc + c^2)$, 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"

WLOG, we can assume that $a\geq b\geq c$. getting that $\frac{1}{a}\leq \frac{1}{b} \leq \frac{1}{c}; a(b+c-a)\leq b(a+c-b)\leq c(a+b-c)$

Applying rearrangement inequality to it, getting that $\frac{a(b+c-a)}{c}+\frac{b(a+c-b)}{a}+\frac{c(a+b-c)}{b}\leq \frac{a(b+c-a)}{a}+\frac{b(a+c-b)}{b}+\frac{c(a+b-c)}{c}=a+b+c$, which equivalent to $\frac{a(b-a)}{c}+\frac{b(c-b)}{a}+\frac{c(a-c)}{b} \leq 0$ Time $abc$ at both side and get the desired inequality ~bluesoul

See Also

1983 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Problem
All IMO Problems and Solutions
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