Nelson Faustino guessed that $(a^2+b^2)(b^2+c^2)(c^2+a^2)\le 8$. Let us prove it.
Proof: We will use the following lemma whose proof is given later.
Lemma 1: Let $a, b, c\ge 0$ with $11(a^2+b^2+c^2) + 4(a+b+c) - 45 \le 0$. Then
$(a^2+b^2)(b^2+c^2)(c^2+a^2)\le 8$.
Let us begin. Using AM-GM, we have
$$\frac{1}{(a^2+b^2)^2} + \frac{1}{(b^2+c^2)^2} + \frac{1}{(c^2+a^2)^2}
\ge 3\sqrt[3]{\frac{1}{(a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2}}.$$
It suffices to prove that $(a^2+b^2)(b^2+c^2)(c^2+a^2)\le 8$.
To this end, we have
\begin{align}
&a, b, c > 0, \quad a^ab^bc^c=1 \qquad\qquad (1)\\
\Longrightarrow \quad & a\ln a + b\ln b + c\ln c = 0, \quad 0 < a, b, c < \frac{159}{100}\qquad\qquad (2)\\
\Longrightarrow \quad &11(a^2+b^2+c^2) + 4(a+b+c) - 45 \le 0, \quad a, b, c > 0.
\qquad\qquad (3)
\end{align}
Here, in (2), since $-u\ln u \le \frac{1}{\mathrm{e}}$ for $u>0$, we have
$a\ln a, b\ln b, c\ln c \le \frac{2}{\mathrm{e}}$ which, when combined with $\frac{159}{100}\ln \frac{159}{100} > \frac{2}{\mathrm{e}}$,
results in $a, b, c < \frac{159}{100}$ (noting that $x \mapsto x\ln x$ is strictly increasing for $x > 1$);
In (3), we have used the fact that
$$x\ln(x) \ge \frac{11}{26}x^2 + \frac{2}{13}x - \frac{15}{26}, \quad \forall 0 < x < \frac{159}{100}.\qquad (4)$$
Then, according to Lemma 1, the desired result follows. We are done.
Proof of (4): It suffices to prove that
$$\ln x \ge \frac{\frac{11}{26}x^2 + \frac{2}{13}x - \frac{15}{26}}{x}, \quad \forall 0 < x < \frac{159}{100}.$$
Let $h(x) = \ln x - \frac{\frac{11}{26}x^2 + \frac{2}{13}x - \frac{15}{26}}{x}$.
We have $h'(x) = -\frac{11(x-1)(x-\frac{15}{11})}{26x^2}$.
Thus, $h(x)$ is decreasing on $(0, 1)$, increasing on $(1, \frac{15}{11})$ and decreasing on $(\frac{15}{11}, \infty)$.
Note that $h(1) = 0$ and $h(\frac{159}{100}) > 0 $. The desired result follows.
Proof of Lemma 1: One of the methods is the uvw method as follows. However, I hope to see nice proofs of Lemma 1.
Let $a+b+c = 3u$, $ab+bc+ca = 3v^2$ and $abc = w^3$.
The constraint $11(a^2+b^2+c^2) + 4(a+b+c) - 45 \le 0$ becomes $11(9u^2 - 6v^2) + 12u - 45 \le 0$.
The constraint does not involve $w^3$.
We need to prove that $f(w^3) = 8 - (a^2+b^2)(b^2+c^2)(c^2+a^2) = w^6+6u(9u^2-6v^2)w^3-81u^2v^4+54v^6+8 \ge 0$.
Since $9u^2 - 6v^2 \ge 0$ and $w^3\ge 0$, $f(w^3)$ is increasing with $w^3$.
Thus, we only need to prove the case when $a=b$ or $c=0$. The rest is not hard and thus omitted.
The conjecture
$$
n\Big(\prod_{i=1}^{n}a_i\Big)^{\frac{1}{n}}\leq n\sqrt{\frac{n}{\sum_{i=1}^{n}\frac{1}{a_i}}}\leq \sum_{i=1}^{n}a_i
$$
holds for $n=2$. For higher $n$, it does not hold, and there are counterexamples where either side of the double inequality is violated.
Here are samples which meet the condition $\sum_{i=1}^{n}a_i\ln(a_i)=0$. Let $n=3$.
The left inequality is violated e.g. for
$a_1 = 0.6, a_2 = 1.3, a_3 = 0.006961$, where upon inserting these numbers the proposed double inequality reads
$0.527282 \le 0.429898 \le 1.906961$.
The right inequality is violated e.g. for
$a_1 = 0.7, a_2 = 1.7, a_3 = 0.275730$, where upon inserting these numbers the proposed double inequality reads
$2.438257 \le 2.750378 \le 2.675730$.
Both samples violate the conjecture. $\qquad \Box$
Best Answer
Even for $f(x)=x^n$ it's not so easy.
We need to prove $$(x+y)\Big(\frac{x^2+y^2}{x+y}\Big)^n(x^n+y^n)\geq 2\left(x^{n+1}+y^{n+1}\right)\Big(\frac{x+y}{2}\Big)^n$$ or $$2^{n-1}(x^2+y^2)^n(x^n+y^n)\geq\left(x^{n+1}+y^{n+1}\right)(x+y)^{2n-1}.$$ Now, let $x=ty$.
Also, since our inequality is symmetric, we can assume that $t\geq1.$
Thus, we need to prove that $g(t)\geq0,$ where $$g(t)=(n-1)\ln2+n\ln(t^2+1)+\ln(t^n+1)-\ln\left(t^{n+1}+1\right)-(2n-1)\ln(t+1).$$ Now, $$g'(t)=\frac{h(t)}{(t^2+1)\left(t^{n+1}+1\right)\left(t^n+1\right)(t+1)},$$ where $$h(t)=n(t-1)^3(t+1)t^{n-1}+2n(t-1)\left(t^{2n+1}+1\right)-(t^2+1)\left(t^{2n}-1\right).$$ Now, prove that $$h(1)=h'(1)=h''(1)=0$$ and $h'''(t)\geq0$ for all $t\geq1.$