0

Given $a,b,c$ are three reals, $a+b+c=2$ and $$\begin{align} (a^2+ab+b^2)(b^2+bc+c^2)&+(b^2+bc+c^2)(c^2+ca+a^2)\\ &+(c^2+ca+a^2)(a^2+ab+b^2)\\ &=625+\frac34(ab+bc+ca)^2 \end{align}$$ Find the value of $|ab+bc+ca|$.

Attempt:

First I tried theory of equations, but it didn’t help. Then I went as follows:

$$a^2+ab+b^2\ge3ab$$ so $$625+\frac{3(ab+bc+ca)^2}{4}\ge9(a^2bc+ab^2c+abc^2) \tag1$$ Taking out $abc$ common, we get $$625+\frac{3(ab+bc+ca)^2}{4}\ge18abc \tag2$$ Writing $(a+b+c)^2=4$ and some manipulations give $$625+\frac{4}{3}\ge18abc \tag3$$

Now I’m stuck here, I’m not sure whether this question can be solved by inequality or not but still this is the best which I could do.

Bill Dubuque
  • 272,048
Maths
  • 495

1 Answers1

3

Note we can express the cyclic sum in LHS in terms of elementary symmetric polynomials, like so: $$\sum_{cyc} (a^2+ab+b^2)(b^2+bc+c^2)=(a+b+c)^4-3(a+b+c)^2(ab+bc+ca)+3(ab+bc+ca)^2$$ Hence using $a+b+c=2$, and letting $t=ab+bc+ca$, we have $$16-12t+3t^2=625+\frac34t^2$$ $$\iff (t+14)(3t-58)=0$$

Now $t=\frac12(a+b+c)^2-\frac12(a^2+b^2+c^2)\leqslant2-\frac12t\implies t\leqslant \frac43$, thus $t=-14$.

P.S. It is not hard to demonstrate there are solutions for $a+b+c=2$, $ab+bc+ca=-14$, for e.g. $\{a, b, c\}=\{1\pm\sqrt{15},0\}$ does the job.

Macavity
  • 46,381
  • Can you please elaborate your approach used to convert the cyclic sum into such a nice expression? – Maths Aug 13 '23 at 12:19
  • Sometimes the approach is easy to notice, like having $a^4$ term on LHS needs $(a+b+c)^4$, and having $a^2b^2,..$ needs $(ab+bc+ca)^2$ etc. However if you want a generic procedure, check Gauss' algorithm referenced in this answer: https://math.stackexchange.com/questions/14051/symmetric-polynomials-and-the-newton-identities/14061#14061 – Macavity Aug 13 '23 at 12:50
  • Thanks a lot for the answer and reference. – Maths Aug 13 '23 at 12:53