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Let $a, b, c$ be the side lengths of a triangle. Prove that $$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right )\geq 3\left ( a+ b- c \right )c\left ( a- b \right )\left ( b- c \right )$$ It can be shown that the constant $3$ is optimal using a discriminant-based method. This inequality is reminiscent of a transformation technique used by Bernhard Leeb in his solution to the IMO 1983 inequality problem, which breaks symmetry by assuming a specific order among the variables.
I want to see an elegant solution for this. Thank you so much for your help.

Dang Dang
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    So is your question that you are looking for a proof of this inequality ? – Hari Ramakrishnan Sudhakar Mar 05 '21 at 04:18
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    Nice! Its quite strong atleast for the case when $a\ge b\ge c$ ,does buffalo way work,? – Hari Ramakrishnan Sudhakar Mar 05 '21 at 04:45
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    @haidangel How about $a = 12, b = 15, c = 17$? – River Li Mar 05 '21 at 05:47
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    By the way, if you change RHS into $24(a+b-c)(a-b)(b-c)c$ then it's true. – NKellira Mar 05 '21 at 06:10
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    @RiverLi I try SOS, but no success, can you? – NKellira Mar 05 '21 at 07:42
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    @tthnew My method: Using Ravi's substitutions, it suffices to prove that $$ \left( 6,y+2,z \right) {x}^{3}+ \left( -6,{y}^{2}-2,yz \right) {x }^{2}+ \left( 2,{y}^{3}-2,{y}^{2}z-8,y{z}^{2} \right) x+6,{y}^{2}{ z}^{2}+2,y{z}^{3} \ge 0. $$ Then prove that the discriminant of cubic (in $x$) is negative or zero. – River Li Mar 05 '21 at 10:26
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    @haidangel Ji Chen posted many interesting and difficult problems in AoPS. You are extracting these problems for us. – River Li Mar 05 '21 at 11:15
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    @haidangel I cannot access that link. But I find that Ji Chen was active in 2020 in that forum. However, the older posts are not there. See: https://bbs.cnool.net/other/ji23 – River Li Mar 06 '21 at 09:14
  • @RiverLi It seems that he doesn't active there from 2021 to now. – NKellira Mar 09 '21 at 13:44
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    @tthnew Yes, but it is about one year. Also, he is the moderator. Please try another way to contact him. Does Sir xzlbq know his address? You can search his paper to see if there is email address. – River Li Mar 09 '21 at 14:00
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    I guess I solved this problem a long time ago. If you need the solution, I can find it from my old notebooks. – lone student Mar 19 '21 at 15:17
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    My solution was related this original inequality $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)≥0$ – lone student Mar 20 '21 at 08:10

1 Answers1

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Just a comment but it's too long for a comment.

Given $a, b, c$ be three side-lengths of a triangle. Prove that $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geqslant 3(a+b-c)(a-b)(b-c)c$$ Source: By MSE/@tthnew

Some try to proof: Since $a,b,c$ be three sid-lengths of a triangle, if $(a-b)(b-c)<0$ it's obvious.

So, we only need to consider this inequality when $(a-b)(b-c)\geq 0\to b\equiv \text{mid}\{a,b,c\}.$

But I can't prove this.

By the way if we change $3\to \frac{12}{5}=2.4,$ it's much easy to prove (for the weaker version).

Indeed, let $a=x+y,b=y+z,c=z+x$ and multiply our inequality with $x+y+z.$ The inequality becomes: $$\frac{1}{20} \left( x-y \right)^2z^2 \left( 15x+31y \right) +\dfrac{1}{20} \left( 2x-y-z \right)^2y \left( 20x^2+19xy+46zx+37 z^2 \right) $$ $$+{\frac {21}{20}} \left( zx+y^2-2yz \right)^2x+\frac {17}{10} \left( x-z \right)^2y^2z+\frac {3}{20} \left( z+x-2y \right) ^2z ^2y$$ $$+\frac{2}{5} \left( x^2-yz \right)^2 \left( 2y+5 z \right) +\frac{1}{5} \left( y-z \right) ^{2}x^3\ge 0.$$

According to Mr. RiverLi's idea, After let $a=x+y,b=y+z,c=z+x$ then $f(a,b,c)=f(x+y,y+z,z+x)\equiv f(x),$ and we only need to prove $$\Delta_x=-16 y^2 (y-z)^4 (y+z) (3 y+z) \left(y^2-4 y z+27 z^2\right)\le 0,$$ which is clearly true. I'm still try to find another way.

NKellira
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