Understanding a technique mentioned in 2016 IMO Shortlist The following question is there in the IMO shortlist 2016:

Let $a, b$ and $c$ be positive real numbers such that min $ (ab, bc, ca) > 1$. Prove that
  $$((a^2 + 1)(b^2 + 1)(c^2 + 1))^{1/3} \leq \left(\frac{a+b+c}{3}\right)^2 + 1$$

In the shortlist document https://www.imo-official.org/problems/IMO2016SL.pdf, after the solution, in the comments section they have stated that the given inequality can also be obtained by first proving the stated lemma and then by "mixing variables".
The stated lemma is as follows:

For any positive real numbers $x, y$ with $xy\geq 1$, we have $$(x^2 + 1)(y^2 + 1)\leq \left(\left(\frac{x + y}{2}\right)^2 + 1\right)^2$$

What does this mean? Is it any specific technique which might prove to be useful in olympiad problems?
 A: 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.

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 $x=a^2$.
A: Because by AM-GM
$$\left(\left(\frac{x+y}{2}\right)^2+1\right)^2-(1+x^2)(1+y^2)=\frac{1}{16}(x-y)^2(x^2+6xy+y^2-8)\geq0.$$
A: Note that $\frac{x+y}2$ is exactly the arithmetic mean of $x,y$. The statement amounts to saying that $f(x)f(y)\geq  f((x+y)/2)^2$, where $f(x)=x^2+1$. Try drawing the graphs $y=f(x)$, $x=f(y)$ on the Cartesian plane, and mark out the lengths $f(x),f(y)$, forming a rectangle. Do the same for $f((x+y)/2)$. This should be sufficient to convince you that one is greater than the other. For a rigorous argument, of course you have to refer to the other answer.
