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Let $a$, $b$ and $c$ be the sides of a triangle. Prove that $$a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c) \leqslant 3abc$$

SOURCE: BANGLADESH MATH OLYMPIAD (Preparatory Question.)

I am very new and novice at this problem. I did a little try but couldn't succeed because I was unable to substitute the left term of the inequality into formula. I know a formula that

$a^\text{3} + b^\text{3} + c^\text{3} - 3abc$ = $(a+b+c)(a^\text{2} + b^\text{2} + c^\text{2} - ab -bc - ca)$. But how to use this formula in that case isn't known to me. And how to show the relation of the both side and when they will become equal?

A small help will be enough for me. Thanks in advance.

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marked as duplicate by Arnaud D., YuiTo Cheng, Jendrik Stelzner, metamorphy, Yanior Weg Jun 6 at 17:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ This is true for any non-negative numbers $a,b,c$ which do not have to side of triangles. Look up this for reference. $\endgroup$ – dezdichado Feb 18 at 17:21
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    $\begingroup$ In the limiting case $a+b=c$ this becomes $2\leq3.$ I wonder if that's supposed to be a $2$ on the right-hand side. $\endgroup$ – saulspatz Feb 18 at 17:25
  • $\begingroup$ @dezdichado How did you derive the OPs inequality from Schur's inequality as formulated in the link? $\endgroup$ – Angela Richardson Feb 18 at 17:26
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    $\begingroup$ If I haven't made a mistake, when a,b,c are the sides of a triangle it can be shown that $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)>2abc$ The two sides are equal when $a+b=c$ and I differentiated the difference w.r.t $c$ assuming $a+b<c.$ $\endgroup$ – saulspatz Feb 18 at 17:47
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    $\begingroup$ @AngelaRichardson just rearrange the inequality and use Schur for when $n = 1.$ $\endgroup$ – dezdichado Feb 18 at 19:36
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Using the Ravi substitution $$a=y+z,b=x+z,c=x+y$$ we have to prove that $$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2-6xyz\geq 0$$ But this is AM-GM:

$$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2\geq 6\sqrt[6]{x^6y^6z^6}=6xyz$$

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  • $\begingroup$ @Sonnhard I don't know about the AM-GM. Would you please explain? Then it will be easy for me to clearify the point. $\endgroup$ – Anirban Niloy Feb 18 at 17:30
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    $\begingroup$ One only needs two variable AM/GM: $x^2y+yz^2\ge 2xyz$ etc. $\endgroup$ – Lord Shark the Unknown Feb 18 at 17:49
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    $\begingroup$ I use that $$\frac{a+b+c+d+e+f}{6}\geq \sqrt[6]{abcdef}$$ for $$a,b,c,d,e,f\geq 0$$ $\endgroup$ – Dr. Sonnhard Graubner Feb 18 at 18:10
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It's true for all non-negatives $a$, $b$ and $c$.

Indeed, since our inequality is symmetric, we can assume that $a\geq b\geq c$ and we obtain: $$3abc-\sum_{cyc}a^2(b+c-a)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$ $$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0.$$

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