Characterizing all triples $(a,b,c)$ of positive reals such that $a^2-ab+bc = b^2-bc+ca = c^2-ca+ab$

Question

Characterize all triples $(a,b,c)$ of positive real numbers such that $$a^2-ab+bc = b^2-bc+ca = c^2-ca+ab$$

This is the equality case of the so-called Vasc inequality. I think the answer is that $a=b=c$ or $a:b:c = \sin^2(4\pi/7) : \sin^2(2\pi/7) : \sin^2(\pi/7)$ and cyclic equality cases. I'm mostly interested if there is a way to derive this reasonably, other than just magically guessing the solutions and then doing degree-counting. This was left as a "good exercise" in Mildorf that has bothered me for years.

Relevant information.

Vasc's inequality here: $(a^2 + b^2 + c^2)^2 \ge 3(a^3b + b^3c + c^3a)$ for all $a, b, c \in \mathbb{R}$.

We have the following identity $$(a^2 + b^2 + c^2)^2 - 3(a^3b + b^3c + c^3a) = \frac{ (x - y)^2 + (y - z)^2 + (z - x)^2}{2}$$ where $x = a^2 - ab + bc, y = b^2 - bc + ca, z = c^2 - ca + ab$. Thus, the equality case is that $x = y = z$, i.e. $a^2 - ab + bc = b^2 - bc + ca = c^2 - ca + ab$.

Answer 1

Too long for comment

Using $\sin(\pi-A)=+\sin A,\cos(\pi-B)=-\cos B,\sin2C=2\sin C\cos C$

$$\dfrac{\sin^2\dfrac{2\pi}7}{\sin^2\dfrac{\pi}7}=4\cos^2\dfrac\pi7=2+2\cos\dfrac{2\pi}7$$

$$\dfrac{\sin^2\dfrac{\pi}7}{\sin^2\dfrac{3\pi}7}=\dfrac{\sin^2\dfrac{6\pi}7}{\sin^2\dfrac{3\pi}7}=4\cos^2\dfrac{3\pi}7=2+2\cos\dfrac{6\pi}7$$

$$\dfrac{\sin^2\dfrac{3\pi}7}{\sin^2\dfrac{2\pi}7}=\dfrac{\sin^2\dfrac{4\pi}7}{\sin^2\dfrac{2\pi}7}=4\cos^2\dfrac{2\pi}7=2+2\cos\dfrac{4\pi}7$$

From How to solve $8t^3-4t^2-4t+1=0$ and factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$,

the roots of $$x^3+x^2-2x-1=0$$ are $2\cos\dfrac{2\pi}7,2\cos\dfrac{4\pi}7,2\cos\dfrac{6\pi}7$

Replace $x$ with $y-2$ to find $$y^3-5y^2+6y-1=0$$

Answer 2

If $c=0$ then $a^2-ab=b^2=ab$, which gives $a=b=c=0.$

Let $abc\neq0$ and $a=xb$.

Thus, from the first equation we obtain: $$a^2-ab-b^2=(a-2b)c.$$ If $a=2b$ so $a=b=c=0$, which is impossible here.

Thus, $c=\frac{a^2-ab-b^2}{a-2b}$ and from the second equation we obtain: $$b^2+\frac{(a-b)(a^2-ab-b^2)}{a-2b}=\frac{(a^2-ab-b^2)^2}{(a-2b)^2}-\frac{(a^2-ab-b^2)a}{a-2b}+ab$$ or $$1+\frac{(x-1)(x^2-x-1)}{x-2}=\frac{(x^2-x-1)^2}{(x-2)^2}-\frac{(x^2-x-1)x}{x-2}+x$$ or $$(x-1)(x^3-5x^2+6x-1)=0,$$ which gives $x=1$ and $a=b=c$ or $$x^3-5x^2+6x-1=0.$$ Now, easy to show that $\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}$, $\frac{\sin^2\frac{\pi}{7}}{\sin^2\frac{3\pi}{7}}$ and $\frac{\sin^2\frac{3\pi}{7}}{\sin^2\frac{2\pi}{7}}$ they are roots of the last equation.

For example: $$\left(\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}\right)^3-5\left(\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}\right)^2+6\cdot\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}-1=$$ $$=\left(4\cos^2\frac{\pi}{7}\right)^3-5\left(4\cos^2\frac{\pi}{7}\right)^2+6\left(4\cos^2\frac{\pi}{7}\right)-1=$$ $$=\left(2+2\cos\frac{2\pi}{7}\right)^3-5\left(2+2\cos\frac{2\pi}{7}\right)^2+6\left(2+2\cos\frac{2\pi}{7}\right)-1=$$ $$=8\cos^3\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-4\cos\frac{2\pi}{7}-1=$$ $$=2\left(4\cos^3\frac{2\pi}{7}-3\cos\frac{2\pi}{7}\right)+6\cos\frac{2\pi}{7}+2+2\cos\frac{4\pi}{7}-4\cos\frac{2\pi}{7}-1=$$ $$=2\cos\frac{2\pi}{7}+2\cos\frac{4\pi}{7}+2\cos\frac{6\pi}{7}+1=$$ $$=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{\sin\frac{\pi}{7}}+1=$$ $$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{\sin\frac{\pi}{7}}+1=0.$$ Since we have no another roots, we are done!

Answer 3

Suppose $a=b$, then $bc=b^2=c^2-cb+b^2$, those $a=b=c$. Which includes cyclic equality cases as well.

Now lets assume WLOG $a>b>c$. Show that in this case above equations don't hold simultaneously.