Understanding a technique mentioned in 2016 IMO Shortlist

I understood the question being about the "mixing variables" technique, not about proving the lemma. While I haven't heard the term, the technique itself is familiar, though I could rarely apply it in an olympiad setting.

Using the lemma on $(a,b), (a,c)$ and $(b,c)$ and multiplying the result we get

$$(a^2+1)^2(b^2+1)^2(c^2+1)^2 \le \left(\left(\frac{a+b}2\right)^2+1\right)^2\left(\left(\frac{a+c}2\right)^2+1\right)^2\left(\left(\frac{b+c}2\right)^2+1\right)^2.$$

Notice that the term on the right has the same form as the term on the left. Let's shorten this with $$f(x,y,z)=(x^2+1)^2(y^2+1)^2(z^2+1)^2.$$ and the above inequality simply becomes

$$f(a,b,c) \le f(\frac{a+b}2, \frac{a+c}2,\frac{b+c}2).$$

Once we've established that $\frac{a+b}2,\frac{a+c}2$ and $\frac{b+c}2$ also fulfill the property that any product of 2 (distinct) of them is at least $1$ (which I'll do below to not disturb the flow of the argument), we can apply the above inequality again, using $a'=\frac{a+b}2,b'=\frac{a+c}2, c'=\frac{b+c}2$:

$$f(a,b,c) \le f(\frac{a+b}2, \frac{a+c}2,\frac{b+c}2) \le f(\frac{b+2a+c}4,\frac{a+2b+c}4,\frac{a+2c+b}4).$$

We can continue to do this, in each step increasing the RHS of the previous inequality by replacing it with function $f$ applied the the pairwise arithmetic mean of the previous arguments.

The important part is that if you take 3 numbers and repeatedly replace them with their pairwise arithmetic mean, you get a series of tuples that converges on their "3-way" arithmetic mean. That can be seen by noticing that the sum of elements for all those tuples stays the same ($a+b+c$) and the difference between highest and lowest element is halved in each step.

So we have a series of inequalities $f(a,b,c) \le f(a_1,b_1,c_1) \le f(a_2,b_2,c_2)\le \ldots$, where $\lim_{n\to\infty}a_n = \lim_{n\to\infty}b_n = \lim_{n\to\infty}c_i = {a+b+c \over 3}$.

Since $f$ is continuous, we finally get

$$f(a,b,c) \le f({a+b+c \over 3},{a+b+c \over 3},{a+b+c \over 3}),$$

which is the statement to prove in the problem, raised to the 6th power.

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Now comes the proof omited above that the "min product condition" is 'inherited' from $(a,b,c)$ to $(\frac{a+b}2,\frac{a+c}2,\frac{b+c}2)$. Since the condition is symmtrical, we can assume $a \le b \le c$ w.l.o.g. We know that $ac,ab \ge 1$. The two smallest among $(\frac{a+b}2,\frac{a+c}2,\frac{b+c}2)$ are $\frac{a+b}2$ and $\frac{a+c}2$, and we get

$$\frac{a+b}2\frac{a+c}2=\frac{a^2+ac+ab+bc}4 \ge \frac{a^2+1+1+\frac1{a^2}}4 \ge 1,$$

using $b,c \ge \frac1a$ and the well known $x+\frac1x \ge 2$ for $x=a^2$.