<s>[INST] 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.
 [/INST] [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).
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