I'm working through a set of inequality problems and I'm stuck on the following question:

Find all sets of solutions for which $$(a^2+b^2+c^2)^2=3(a^3b+b^3c+c^3a)$$ holds. Note that $a,b,c\in\mathbb{R}$.

Firstly, one can easily see that when $a=b=c$ then the equality holds: $$\text{LHS}=(a^2+a^2+a^2)^2=(3a^2)^2=9a^4$$ and $$\text{RHS}=3(a^4+a^4+a^4)=3(3a^4)=9a^4.$$

A hint is given to use the substitutions $a=x+2ty$, $b=y+2tz$ and $c=x+2tz$ for real $t$. The LHS is rather nice in that it simplifies to $$(4t(xy+xz+yz)+(1+4t^2)(x^2+y^2+z^2))^2$$ but I can't find a similar simplification for the RHS. I guess this is a type of uvw question but I don't know where to start.

There are actually four sets of solutions: $$a=b=c,$$ $$\frac{a}{\sin^2\frac{4\pi}7}=\frac{b}{\sin^2\frac{2\pi}7}=\frac{c}{\sin^2\frac{\pi}7},$$ $$\frac{b}{\sin^2\frac{4\pi}7}=\frac{c}{\sin^2\frac{2\pi}7}=\frac{a}{\sin^2\frac{\pi}7},$$ and $$\frac{c}{\sin^2\frac{4\pi}7}=\frac{a}{\sin^2\frac{2\pi}7}=\frac{b}{\sin^2\frac{\pi}7}$$ I have no idea how the trigonometric expressions are obtained. How could the equality be solved?

  • $\begingroup$ i think this is an inequality with $a,b,c>0$ $\endgroup$ – Dr. Sonnhard Graubner Jan 1 '18 at 20:35

The hint.

Let $b=a+u$,$c=a+v$ and $u=xv$.

Hence, $$(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)=\sum_{cyc}(a^4+2a^2b^2-3a^3b)=$$ $$=(u^2-uv+v^2)a^2+(u^3-5u^2v+4uv^2+v^3)a+u^4-3u^3v+2u^2v^2+v^4\geq0$$ because $$(u^3-5u^2v+4uv^2+v^3)^2-4(u^2-uv+v^2)(u^4-3u^3v+2u^2v^2+v^4)=$$ $$=-3(u^3-u^2v-2u^2+v^3)^2=-3v^6(x^3-x^2-2x+1)^2\leq0.$$ The equality occurs for $x^3-x^2-2x+1=0$, which gives $$x\in\left\{2\cos\frac{\pi}{7},-2\cos\frac{2\pi}{7},2\cos\frac{3\pi}{7}\right\}.$$ For example, $x=2\cos\frac{\pi}{7}$ gives the following case. $$\frac{b}{\sin^2\frac{4\pi}7}=\frac{c}{\sin^2\frac{2\pi}7}=\frac{a}{\sin^2\frac{\pi}7}.$$ My solution by $uvw$ see here: https://artofproblemsolving.com/community/c6h6026p5329091

There is also the following nice solution. https://gbas2010.wordpress.com/2010/01/08/problem-19vasile-cirtoaje/


Take the square root of all terms in $\frac{c}{\sin^2\frac{4\pi}7}=\frac{a}{\sin^2\frac{2\pi}7}=\frac{b}{\sin^2\frac{\pi}7}$.
Now you obtain a case of a well known equation: the sine rule.
$\frac {4\pi}7 + \frac {2\pi}7 +\frac {\pi}7= \pi$, so this is a triangle with side lengths $\sqrt a, \sqrt b$ and $\sqrt c$ and angles $\frac {\pi}7, \frac {2\pi}7 $ and $ \frac {4\pi}7$.

Hope you can take it from here!

  • $\begingroup$ But $a$, $b$ and $c$ are reals and why they are all solutions? $\endgroup$ – Michael Rozenberg Jan 1 '18 at 20:42
  • $\begingroup$ See Dr. Sonnhard Graubner's comment. $\endgroup$ – Mohammad Zuhair Khan Jan 2 '18 at 7:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.