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
$$\begin{align}
f(x) &= (x-a)(x-b)(x-c)(x-d) \\[4pt]
&= x^4 - 28x^3 + \frac{82}{3}x^2 - (abc+abd+acd+bcd)x + 1
\end{align}$$
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.
 A: 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 $0<a\leq b\leq c\leq d$.
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 $$abc+abd+acd+bcd=\left(\frac{x^4-28x^3+\frac{82}{3}x^2+1}{x}\right)_{x=\frac{1}{3}}=\frac{244}{27}.$$
A: [Here's an unsatisfactory answer as I don't expect the discriminant to be calculated in a contest-math setting.]
Let $ C = abc + bcd + cda + dab > 0$.
The equation $ x^4 - 28x^3 + \frac{82}{3} x^2 - Cx + 1 =0 $ has 4 positive real roots.
The discriminant (Thanks Wolfram) is
$$\Delta = -5565901568/81 + (82093312 C)/9 + (13588640 C^2)/27 - 74032 C^3 - 27 C^4 \\ 
= -9 ( C - \frac{244}{27} ) ^2(3C^2 + 8290C + 93488). $$
Since the equation has 4 real roots, the discriminant is non-negative.
The only positive value of $c$ which makes $\Delta$ non-negative is $ c = \frac{244}{27}$.
Hence, the solution is uniquely determined (up to permutation).
A: Assume $d = \max{a,b,c,d}$. Looking at the inequality:
$$(a+b+c)^2\geq 3(ab+bc+ca)$$
beginning edit by Will:
from Michael,
$$ 82 = 3 (bc+ca+ab) + 3d(a+b+c),  $$
from displayed inequality
$$  82 \leq (a+b+c)^2 + 3d(a+b+c) $$
$$ 82 \leq (28-d)^2 + 3 d (28-d) $$
$$ 82 \leq 784 - 56d + d^2 + 84d - 3 d^2   $$
$$ 0 \leq 702 + 28 d - 2 d^2 $$
$$ 0 \geq 2 d^2 - 28 d - 702  $$
$$  0 \geq d^2 + 14 d - 351 $$
$$ 0 \geq (d+13)(d-27).  $$
As $d >0$ we get
$$ 0 \geq  d-27 $$
$$ 27 \geq d $$
end of edit by Will
will give you $d\leq 27.$ Consequently, $abc\geq \dfrac{1}{27}.$
SECOND EDIT by WILL
$$ f = ( ab + bc + ca)^2 - 3abc(a+b+c) $$
$$ 4(b^2 - bc + c^2)  f =  \left( 2 (b^2 - bc + c^2)  a - bc(b+c)    \right)^2 + 3b^2  c^2 (b-c)^2 $$
Conclusion: permute the letters, $ f \geq 0$ and $f \neq 0$ unless $a=b=c.$ Real $a,b,c$ otherwise unrestricted
END SECOND EDIT by WILL
From $a+b+c \geq 1$and $abc\geq \dfrac{1}{27},$  we find that $ab+bc+ca\geq \dfrac{1}{3}.$ Then,
$$\dfrac{1}{3}\leq ab+bc+ca = \dfrac{82}{3} - d(28-d)\iff d^2-28d+27 \geq 0$$
This means $(d-27)(d-1)\geq 0$ so $d = 27.$ The rest should follow immediately.
A: 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.
