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Relevant wiki: Jensen's Inequality

Define $f(x)=1-\dfrac{1}{1+x}$ and notice that the given inequality is similar to $\displaystyle \sum_{cyclic} \left(1-\dfrac{1}{1+x}\right)\le 1$ . Now since given quantities are positive we have $f''(x)\gt 0$ which implies $f$ is convex throughout the domain. Thus if we use jensen's inequality,

$\displaystyle f(x)+f(y)+f(z)+f(t)\ge 4f\left(\dfrac{x+y+z+t}{4}\right)=4-\dfrac{4}{1+\dfrac{x+y+z+t}{4}}$ , Now if $\alpha = xyzt$ then we can deduce by AM-GM that $\dfrac{x+y+z+t}{4}\ge \sqrt[4]{\alpha}$ and thus $\displaystyle -\dfrac{1}{1+\dfrac{x+y+z+t}{4}} \ge -\dfrac{1}{\sqrt[4]{\alpha}+1}$ .

This hence reduces to $\displaystyle 1\ge \sum_{cyclic} \left(1-\dfrac{1}{1+x}\right) \ge 4-\dfrac{4}{1+\sqrt[4]{\alpha}}$ , on solving we have $\alpha=xyzt\le \dfrac{1}{81}$ making the answer $\boxed{82}$ .

Applying the AM-GM inequality, we have $\dfrac{1}{1+x}=\left(1-\dfrac{1}{1+y}\right)+\left(1-\dfrac{1}{1+z}\right)+\left(1-\dfrac{1}{1+t}\right)=\dfrac{y}{1+y}+\dfrac{z}{1+z}+\dfrac{t}{1+t}\ge3\sqrt[3]{\dfrac{yzt}{(1+y)(1+z)(1+t)}}$ Similarly, $\dfrac{1}{1+y}\ge3\sqrt[3]{\dfrac{xzt}{(1+x)(1+z)(1+t)}};\dfrac{1}{1+z}\ge3\sqrt[3]{\dfrac{xyt}{(1+x)(1+y)(1+t)}};\dfrac{1}{1+t}\ge3\sqrt[3]{\dfrac{xyz}{(1+x)(1+y)(1+z)}}$

Multiple side by side all above inequalities, we get $\dfrac{1}{(1+x)(1+y)(1+z)(1+t)}\ge\dfrac{81xyzt}{(1+x)(1+y)(1+z)(1+t)}$

Or $xyzt\le\dfrac{1}{81}$ .

The equality holds iff $x=y=z=t=\dfrac{1}{3}$ .

We can maximize this equation as we suppose WLOG, that x=y=z=t.

We observe that,

4/1+x >= 3

This gives us

x >= 1/3

So the value of xyzt = x^4

So, if we take the original equation, and raise both sides to the power of 4, we get.

x^4 >= 1/81 where g.c.d(1, 81) = 1

Hence, gives us the answer 1+81 = 82

Using symmetry arguments we can say that at maximum $xyzt, x = y = z = t$ . This means $x <= \frac{1}{3}$ . Hence maximum $xyzt = \frac{1}{81}$ and hence the answer is 82.