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In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle

$ABC$ is equilateral

$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$

$\displaystyle 2\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right) = \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)+3$

let $\displaystyle \frac{a}{b} = x,\frac{b}{c}=y,\frac{c}{a}=z$

so $\displaystyle 2(x+y+z) = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3$

wan,t be able to proceed after that, could some help me

Michael Rozenberg
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DXT
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3 Answers3

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Since $a,b,c$ are sides of a triangle we may write $(a,b,c) = (y+z,z+x,x+y)$ with $x,y,z>0$. The difference between the two sides is then $$ x^3 + y^3 + z^3 \; + \; x^2 y + y^2 z + z^2 x - 2 (xy^2 + yz^2 + zx^2). $$ By cyclic symmetry we may assume $x \geq z$ and $y \geq z$, and write $x = (1 + \alpha) z$ and $y = (1 + \beta) z$ with $\alpha,\beta \geq 0$ to find $z^3$ times $$ 2(\alpha^2 - \alpha\beta + \beta^2) + \beta(\alpha-\beta)^2 + \alpha^3, $$ in which each term is nonnegative, and all are zero iff $\alpha = \beta = 0$ $-$ which in turn is equivalent to $x=y=z$, and thus $a=b=c$, so triangle $ABC$ is equilateral as claimed. $\Box$

[Added later: The solution above applies a general technique for such problems; but I see that for the present question Hari Shankar posted (in a comment) a link to an AoPS item that gives an even simpler conclusion that retains the cyclic symmetry: $$ x^3 + y^3 + z^3 \; + \; x^2 y + y^2 z + z^2 x - 2 (xy^2 + yz^2 + zx^2) = x(x-z)^2 + y(y-x)^2 + z (z-y)^2, $$ again with all terms nonnegative; so equality implies that they all vanish and the rest follows as before. That link also gives the example $(a:b:c) = (10:3:24)$ of a rational point on the cubic $2(bc^2 + ca^2 + ab^2) = b^2c + c^2a + a^2b + 3abc$ with positive variables that do not satisfy the triangle inequality.]

Noam D. Elkies
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  • I think, your solution is very ugly. See my solution. – Michael Rozenberg Jan 08 '17 at 05:34
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    Your solution is essentially the same as the AoPS solution that I quote at the end of my answer (except that you kept everything in terms of the original $a,b,c$, making the displayed formulas longer). That's prettier than the first solution I gave (though surely it's not all that ugly...), but this solution illustrates a general technique, and thus does not attempt to exploit simplifications available for this particular instance of this class of problems. – Noam D. Elkies Jan 08 '17 at 05:43
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Because $$0=\sum\limits_{cyc}(2b^2a-a^2b-abc)=\sum\limits_{cyc}(a^3-abc-(b^3-2b^2a+a^2b))=$$ $$=(a+b+c)\sum\limits_{cyc}(a^2-ab)-\sum\limits_{cyc}b(a-b)^2=$$ $$=\sum\limits_{cyc}(a-b)^2\left(\frac{1}{2}(a+b+c)-b\right)=\frac{1}{2}\sum\limits_{cyc}(a-b)^2(a+c-b).$$ Id est, the given it's $$(a-b)^2(a+c-b)+(b-c)^2(b+a-c)+(c-a)^2(c+b-a)=0,$$ which says that $a=b=c$.

Done!

Michael Rozenberg
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$x+y+z\geq 3(xyz)^{\frac{1}{3}}$,when $x=y=z$ ,equal sign is established.So:$$2\sum\frac{c}{a}\geq\sum\frac{c}{a}+3$$ i.e.$$\frac{c}{a}+\frac{b}{c}+\frac{a}{b}\leq\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$$Without loss of generality, we assume:$c\geq b\geq a$,so $\frac1{a}\geq\frac1{b}\geq\frac1{c},$then we have:$$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\leq\frac{c}{a}+\frac{b}{c}+\frac{a}{b},$$so $a=b=c$.I'm using an incomplete sort to derive a contradiction.

Zuo
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  • Note that the RHS sum is $\sum\dfrac{a}{c}$. – Darth Geek Dec 29 '16 at 18:44
  • I will make it clearer – Zuo Dec 29 '16 at 18:53
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    but Zeo RHS present $\sum \frac{a}{c}$ and you wrote $\sum \frac{c}{a}$ – DXT Dec 30 '16 at 04:59
  • Since neither the original equality nor your inequality is symmetric on $a,b,c$ you can't assume WLOG that $c\geq b\geq a$. Also, you prove that $\sum \dfrac{c}{a} \geq 3$, not that $2\sum \dfrac{c}{a} \geq \sum \dfrac{a}{c} + 3$. – Darth Geek Dec 30 '16 at 09:26
  • @Darth Geek I think you should check it again.I prove that$B+3=2A\geqA+3$,so,$B\geqA$ – Zuo Dec 30 '16 at 11:49
  • @Zuo $x+y+z \geq 3(xyz)^{1/3}$ proves that $x+y+z \geq 3$, i.e. $\sum \dfrac{c}{a} \geq 3$. With equality iff $a = b = c$. That doesn't prove that $2\sum \dfrac{c}{a} \geq \sum \dfrac{a}{c} + 3$. Incidently, you still write $\sum \dfrac{c}{a}$ on the RHS, where it's supposed to be $\sum \dfrac{a}{c}$ instead. – Darth Geek Dec 30 '16 at 11:56
  • Please look carefully – Zuo Dec 30 '16 at 12:21
  • Right, I see what you are trying to do now. You might want to add some more text so it's clearer. However, since the inequality you try to prove is not symmetric on $a,b,c$ you can't assume $c\geq b\geq a$ WLOG. In fact, if $b \leq c \leq a$ then you should get the opposite inequality: $$\dfrac{a}{c} + \dfrac{b}{a} + \dfrac{c}{b} \geq \dfrac{c}{a} + \dfrac{b}{c} + \dfrac{a}{b}$$ – Darth Geek Dec 30 '16 at 12:58
  • Also, if $a = b < c$ then indeed $c\geq b\geq a$ and $$\dfrac{a}{c} + \dfrac{b}{a} + \dfrac{c}{b} =\dfrac{a}{c} + 1 + \dfrac{c}{a} = \dfrac{c}{a} + \dfrac{b}{c} + \dfrac{a}{b}$$ but you can't deduce from that that $a = b = c$. – Darth Geek Dec 30 '16 at 13:11