Suppose $a,b,c,p,q,r$ are some positive real parameters, none of which exceeds the sum of the other five. Consider a real-valued function of a real variable $x$ defined as $$\!\!\!\!\small{\begin{align}f(x)&=\sqrt{(a+b+c-x) (x-a+b+c) (x+a-b+c) (x+a+b-c)}\\&+\unicode{x200a}\sqrt{(p+q+r-x) (x-p+q+r) (x+p-q+r) (x+p+q-r)}.\end{align}}\tag{$\small\diamondsuit$}$$ We assume that it is only defined for positive values of $x$ such that both expressions under the square roots are also positive. We are interested in the largest value attained by $f(x)$. It is possible to prove that it happens when $x$ is one of the roots of the septic equation $$\!\!\!\!\small{\begin{align}&(a+b+c-x)(x-a+b+c)(x+a-b+c)(x+a+b-c)(p\, x+q\, r)(q\, x+p\, r)(r\, x+p\, q)=\\&(p+q+r-x)(x-p+q+r)(x+p-q+r)(x+p+q-r)(a\, x+b\, c)(b\, x+a\, c)(c\, x+a\, b).\end{align}}\tag{$\small\spadesuit$}$$ It is obvious that the largest value of $f(x)$ is the same for all permutations of the triples $(a,b,c)$ and $(p,q,r)$ because the parameters within each triple appear in the definition of $f(x)$ in a symmetrical way.
But it looks like the largest value actually remains the same for all permutations of the sextuple $(a,b,c,p,q,r)$, although the value of $x$ that maximizes $f(x)$ can be different for different permutations. This property seems to not be immediately obvious from the definition. How can we prove it?