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Once again I have come across an olympiad-type problem which probably requires some sort of insight even though it looks simple. The question is as follows:

Let $a$, $b$ and $c$ be positive real numbers. Prove that:

$(a+b)(b+c)(c+a)$ $\geqslant$ $8(a+b-c)(b+c-a)(c+a-b)$

I have tried to multiply out the LHS but unfortunately it didn't get me much...

I found that if one of $a$, $b$ or $c$ is greater than or equal to the sum of the other two, then the inequality is trivially true, since LHS is positive while RHS isn't.

Would there be a quick and easy formula or known inequality that I could use to make this problem simpler? Or is this just a 'bash-and-solve' type question?

Any help, comments or edits are greatly appreciated! Thanks! :)

This question appeared in the South African Mathematics Olympiad in 2008.

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5 Answers 5

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Hint: We may make the assumption that each of the terms on the RHS are positive.

Hint: Use the substitution $$ x = a+b - c \\ y = b+c -a \\ z = c+a - b \\$$

What happens now?

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By homogeneity, we may assume wlog that $a + b + c = 1$. We want to minimize $f(a,b,c) = \left( a+b \right) \left( b+c \right) \left( c+a \right) -8\, \left( a+b-c \right) \left( b+c-a \right) \left( c+a-b \right)$ on the triangle $a+b+c=1$, $a,b,c\ge 0$. Critical points with $a,b,c>0$ are found using a Lagrange multiplier: I get $(1/3,1/3,1/3)$ with $f(1/3,1/3,1/3) = 0$ and $(31/63, 31/63, 1/63)$ and its permutations with $f(31/63, 31/63, 1/63) = 1000/3969$. We must also look at the boundary, but I find that $f(a,b,0) = (a+b)(8 a^2 - 15 a b + 8 b^2) \ge 0$ for $a,b\ge 0$.

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  • $\begingroup$ :hi .why do you assume a + b + c = 1? $\endgroup$
    – M.H
    Mar 19, 2013 at 21:25
  • $\begingroup$ @MaisamHedyelloo if not, divide all of $\{a,b,c\}$ by the sum and it cancels out from the original inequality $\endgroup$
    – gt6989b
    Mar 19, 2013 at 21:43
  • $\begingroup$ Because all terms are homogeneous of degree $3$, i.e. change $a,b,c$ to $ta, tb, tc$ and you multiply everything by $t^3$. $\endgroup$ Mar 19, 2013 at 22:50
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Download the solution of the 2008 question paper from the SA Mathemataics Foundation website here - http://www.samf.ac.za/QuestionPapers.aspx

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  • $\begingroup$ Oh thanks, when I tried to access the answers previously the page didn't load for some reason, but now it does :| Thanks! :) $\endgroup$
    – mathsnoob
    Mar 26, 2013 at 6:32
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Since our inequality is symmetric, it's enough to assume that $a\geq b\geq c$, which gives: $$\prod_{cyc}(a+b)-8\prod_{cyc}(a+b-c)=$$ $$=\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)-8\sum_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)=$$ $$=\sum_{cyc}(8a^3-7a^2b-7a^2c+6abc)=$$ $$=\sum_{cyc}(8a^3-4a^2b-4a^2c-(3a^2b+3a^2c-6abc))=$$ $$=\sum_{cyc}4(a-b)^2(a+b)-3\sum_{cyc}c(a-b)^2=\sum_{cyc}(a-b)^2(4a+4b-3c)\geq$$ $$\geq(a-c)^2(4a+4c-3b)+(b-c)^2(4b+4c-3a)\geq$$ $$\geq(b-c)^2(4a+4c-3b)+(b-c)^2(4b+4c-3a)=(b-c)^2(a+b+8c)\geq0.$$

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Given $\quad(a+b)(b+c)(c+a) \ge 8(a+b-c)(b+c-a)(c+a-b)\quad$ show $x\ge y$ $$x=(a+b)(b+c)(c+a) \quad=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2$$ $$y=8(a+b-c)(b+c-a)(c+a-b) \\ = -8 a^3 + 8 a^2 b + 8 a^2 c + 8 a b^2 - 16 a b c + 8 a c^2 - 8 b^3 + 8 b^2 c + 8 b c^2 - 8 c^3$$

We negate the RHS and subtract it from both sides by addition so $x-y \ge 0$

$$x-y\quad =\quad a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2\\ +8 a^3 - 8 a^2 b - 8 a^2 c - 8 a b^2 + 16 a b c - 8 a c^2 + 8 b^3 - 8 b^2 c - 8 b c^2 + 8 c^3\\ =8a^3-7a^2b-7a^2c-7ab^2+18abc-7ac^2+8b^3-7b^2c-7bc^2+8c^3\quad \ge\quad 0\\ \implies 8(a^3+b^3+c^3)\quad\ge\quad 7(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2)$$ If $\quad a=b=c=1\quad$ then $\quad 24\ge 42\quad $which is a contradiction.

If $\quad a=b=c=2\quad$ then $\quad 192 \ge 336\quad $ which is a contradiction.

If $\quad a=b=c=3\quad$ then $\quad 648 \ge 1134\quad $ which is a contradiction.

This math could be wrong but it appears that the original statement is reversed.

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  • $\begingroup$ you forgot about 18abc. $\endgroup$
    – Bek
    Nov 2, 2021 at 7:57

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