1983 IMO Problems/Problem 6

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

1983 IMO (Problems) • Resources
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