Solving for positive reals: $abcd=1$, $a+b+c+d=28$, $ac+bc+cd+da+ac+bd=82/3$ $$a,b,c,d \in \mathbb{R}^{+}$$
$$ a+b+c+d=28$$
$$ ab+bc+cd+da+ac+bd=\frac{82}{3} $$
$$ abcd = 1 $$
One can also look for the roots of polynomial
$$\begin{align}
f(x) &amp;= (x-a)(x-b)(x-c)(x-d) \\[4pt]
&amp;= x^4 - 28x^3 + \frac{82}{3}x^2 - (abc+abd+acd+bcd)x + 1
\end{align}$$
and $f(x)$ has no negative roots... but how else do I proceed?
There is a trivial solution $\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 27$. We just need to prove it's unique.
The hint.
By your work  $$\frac{x^4-28x^3+\frac{82}{3}x^2+1}{x}-\frac{244}{27}=\frac{(3x-1)^3(x-27)}{27x}$$ because by Rolle $$\left(\frac{x^4-28x^3+\frac{82}{3}x^2+1}{x}\right)'=\frac{(3x-1)^2(x^2-18x-3)}{3x^2}$$ has three positive roots and one of them must be $\frac{1}{3},$ which is also an element of $\{a,b,c,d\}$.
Indeed, let $0&lt;a\leq b\leq c\leq d$.
Thus, $f'$ has positive roots on $[a,b]$ on $[b,c]$ and on $[c,d]$ and we know that one of these roots it's double $\frac{1}{3}$.
Let $\frac{1}{3}\in[a,b]$ and $\frac{1}{3}\in[b,c].$
Thus, $b=\frac{1}{3}$, which says $$abc+abd+acd+bcd=\left(\frac{x^4-28x^3+\frac{82}{3}x^2+1}{x}\right)_{x=\frac{1}{3}}=\frac{244}{27}.$$
Use Mathematica to directly calculate this equation system, there are 4 sets of solutions that meet the requirements:

The output solution is only for comparison, not a substitute for theoretical analysis.