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For any triangle with sides $a$, $b$, $c$, prove the inequality $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0 .$$

This is IMO 1983 problem 6.

I tried substituting $a=x+y$, $b=y+z$, $c=z+x$ but well it doesn't help in any sense except wasting 3 pages that lead to nothing (please don't mind the joke). Using $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$ also didn't lead to anything for me. Could you give me a hint for finding the proper substitution?

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  • $\begingroup$ Proper substitution for what? $\endgroup$ – Jessica B Dec 28 '14 at 7:53
  • $\begingroup$ obviously for a,b,c so that the inequality can be simplified to a nicer and more "do-able" inequality.. $\endgroup$ – Abhishek Bakshi Dec 28 '14 at 7:54
  • $\begingroup$ What does it mean to 'do' an inequality? An inequality is a statement, not an activity. $\endgroup$ – Jessica B Dec 28 '14 at 8:00
  • $\begingroup$ The triangle's incircle divide the edges into $a=y+z,b=x+z, c=x+y$ for some positive numbers $x,y,z$. $\endgroup$ – Empy2 May 26 at 10:59
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The Ravi substitution that you did works just fine.

With $a = y+z,b = x+z$ and $c = x+y$, we will need to prove:

$$\sum\limits_{cyc} (y+z)^2(z+x)(y-x) \ge 0$$

Simplifies to, $xy^3 + yz^3 + zx^3 \ge x^2yz + xy^2z + xyz^2$

Use Cauchy-Schwarz Inequality:

$\displaystyle \begin{align}x^2yz + xy^2z + xyz^2 &= \sum\limits_{cyc} (x^{3/2}z^{1/2})(x^{1/2}yz^{1/2}) \\&\le (x^3z + y^3x + z^3y)^{1/2}(xy^2z + xyz^2 + x^2yz)^{1/2}\end{align}$

Thus, $x^2yz + xy^2z + xyz^2 \le xy^3 + yz^3 + zx^3$

The link provided by @math110 gives a simpler explanation:

Rewrite the inequality as:

$\displaystyle \sum\limits_{cyc} \frac{y^2}{z} \ge x+y+z$ which is Engel's form $\displaystyle \left(\sum\limits_{cyc} \frac{y^2}{z}\right)(x+y+z) \ge (x+y+z)^2$

or we can use Rearrangement as @math110 suggests.

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    $\begingroup$ Nice,+1 This is IMO 1983 problem,artofproblemsolving.com/Forum/… $\endgroup$ – math110 Dec 28 '14 at 8:06
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    $\begingroup$ @math110 ah ! Thanks :D oops ! looks like I over complicated things with the CS ineq. $\endgroup$ – r9m Dec 28 '14 at 8:08
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    $\begingroup$ so It's Nice,and can use rearrangement inequality. $\endgroup$ – math110 Dec 28 '14 at 8:10
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Let $c=\max\{a,b,c\}$, $a=x+u$, $b=x+v$ and $c=x+u+v$, where $x>0$ and $u\geq0$, $v\geq0$.

Hence, $$\sum_{cyc}(a^3b-a^2b^2)=(u^2-uv+v^2)x^2+(u^3+2u^2v-uv^2+v^3)x+2u^3v\geq0.$$ Done!

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Substitution $$a=y+z,b=x+z,c=x+y$$ we have to prove that $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z$$ Using Cauchy Schwarz in Engel form we get $$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq \frac{(x+y+z)^2}{x+y+z}=x+y+z$$

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$$\left \{ \sum\limits_{cyc}\,a^{\,2}b\,(\,a- b\,) \right \}= \frac{bc(\,a+ b- c\,)^{\,2}(\,a- b\,)^{\,2}}{b(\,a+ b- c\,)+ 3\,a(\,c+ a- b\,)}+$$ $$+ \dfrac{a(\,c+ a- b\,)\left [ 3\left \{ \sum\limits_{cyc}\,a^{\,2}b\,(\,a- b\,) \right \}- b(\,a+ b- c\,)(\,a- c\,)(\,c- b\,) \right ]}{b(\,a+ b- c\,)+ 3\,a(\,c+ a- b\,)}\geqq 0 \tag{29}$$ We need to $\lceil$ Prove $3\left \{ \sum\limits_{cyc}\,a^{\,2}b\,(\,a- b\,) \right \}\geqq b(\,a+ b- c\,)(\,a- c\,)(\,c- b\,)$ $\rfloor$

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Let $LHS=f(a,b,c)$ and note that $f(a,b,c)\ge f(a-k,b-k,c-k)$ for $k\le min\{a,b,c\}$ so WLOG, we can take $c=0$. Now we have to show $a^2b(a-b)\ge0$. Because, we are working on a triangle $a\le b+c=b$ and $b\le a+c=a$, thus $a=b$. So $a^2b(a-b)\ge0$ is true, we have done!

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