, 2012

Let $a = BC$, $b = AC$, $c = AB$, and $\angle A = \alpha$, $\angle B = \beta$, $\angle C = \gamma$. Let $AH_1$, $BH_2$, $CH_3$ be altitudes, $AL_1$, $BL_2$, $CL_3$ be angle bisectors, and $AM_1$, $BM_2$, $CM_3$ be medians of $\triangle ABC$.
Then
$$\frac{AI}{L_1I} = \frac{b+c}{a}, \quad \frac{BI}{L_2I} = \frac{a+c}{b}, \quad \frac{CI}{L_3I} = \frac{a+b}{c}.$$
Since $AG : GM_1 = BG : GM_2 = CG : GM_3 = 2 : 1$, we have
$$f(G) = \frac{AG}{M_1G} \cdot \frac{BG}{M_2G} \cdot \frac{CG}{M_3G} = 2 \cdot 2 \cdot 2 = 8.$$
Hence,
$$\begin{aligned} f(I) &= \frac{AI}{L_1I} \cdot \frac{BI}{L_2I} \cdot \frac{CI}{L_3I} = \frac{b+c}{a} \cdot \frac{a+c}{b} \cdot \frac{a+b}{c} \ge \frac{2\sqrt{bc}}{a} \cdot \frac{2\sqrt{ac}}{b} \cdot \frac{2\sqrt{ab}}{c} = 8 = f(G), \end{aligned}$$
i.e. $f(I) \ge f(G)$ with the equality if and only if $a = b = c$.

Further,

$$\begin{aligned} HH_1 &= BH_1 \cdot \tg \angle HBH_1 = \\ &= (AB \cos \beta) \cdot \tg (90^\circ - \gamma) = \frac{c \cdot \cos \beta \cdot \cos \gamma}{\sin \gamma} \end{aligned}$$
and
$$AH = \frac{AH_2}{\cos \angle HAH_2} = \frac{AB \cdot \cos \alpha}{\cos(90^\circ - \gamma)} = \frac{c \cdot \cos \alpha}{\sin \gamma}.$$
So we get $\frac{AH}{H_1H} = \frac{\cos \alpha}{\cos \beta \cos \gamma}$. Similarly, $\frac{BH}{H_2H} = \frac{\cos \beta}{\cos \alpha \cos \gamma}$ and $\frac{CH}{H_3H} = \frac{\cos \gamma}{\cos \alpha \cos \beta}$.
$$\text{Therefore, } f(H) = \frac{AH}{H_1H} \cdot \frac{BH}{H_2H} \cdot \frac{CH}{H_3H} = \frac{1}{\cos \alpha \cos \beta \cos \gamma}.$$
Let $m = \operatorname{tg}\frac{\alpha}{2}$, $n = \operatorname{tg}\frac{\beta}{2}$ and $k = \operatorname{tg}\frac{\gamma}{2}$. By condition, $\alpha, \beta, \gamma \in (0^\circ, 90^\circ)$, so $m, n, k \in (0, 1)$. Therefore,
$$f(H) = \frac{1}{\cos \alpha \cos \beta \cos \gamma} = \frac{1+m^2}{1-m^2} \cdot \frac{1+n^2}{1-n^2} \cdot \frac{1+k^2}{1-k^2}$$
We have
$$\begin{aligned} \frac{b+c}{a} &= \frac{b}{a} + \frac{c}{a} = \frac{\sin \beta}{\sin \alpha} + \frac{\sin \gamma}{\sin \alpha} = \frac{2 \sin \frac{\beta+\gamma}{2} \cos \frac{\beta-\gamma}{2}}{\sin(180^\circ - (\beta+\gamma))} = \\ &= \frac{2 \sin \frac{\beta+\gamma}{2} \cos \frac{\beta-\gamma}{2}}{\sin(\beta+\gamma)} = \frac{\cos \frac{\beta-\gamma}{2}}{\cos \frac{\beta+\gamma}{2}} = \frac{\cos \frac{\beta}{2} \cos \frac{\gamma}{2} + \sin \frac{\beta}{2} \sin \frac{\gamma}{2}}{\cos \frac{\beta}{2} \cos \frac{\gamma}{2} - \sin \frac{\beta}{2} \sin \frac{\gamma}{2}} = \\ &= \frac{1 + \operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\gamma}{2}}{1 - \operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\gamma}{2}} = \frac{1 + nk}{1 - nk}. \end{aligned}$$
Similarly, $\frac{a+c}{b} = \frac{1+mk}{1-mk}$ and $\frac{a+b}{c} = \frac{1+mn}{1-mn}$. So,
$$f(I) = \frac{a+b}{c} \cdot \frac{b+c}{a} \cdot \frac{c+a}{b} = \frac{1+mn}{1-mn} \cdot \frac{1+nk}{1-nk} \cdot \frac{1+km}{1-km}$$
Thus,
$$f(H) \ge f(I) \iff \frac{1+m^2}{1-m^2} \cdot \frac{1+n^2}{1-n^2} \cdot \frac{1+k^2}{1-k^2} \ge \frac{1+mn}{1-mn} \cdot \frac{1+nk}{1-nk} \cdot \frac{1+km}{1-km}$$
Lemma. If $x, y \in (0,1)$, then $\frac{1+x^2}{1-x^2} \cdot \frac{1+y^2}{1-y^2} \ge \left(\frac{1+xy}{1-xy}\right)^2$ with the equality if and only if $x=y$.
Proof. We have
$$\begin{aligned} \frac{1+x^2}{1-x^2} \cdot \frac{1+y^2}{1-y^2} &\ge \left(\frac{1+xy}{1-xy}\right)^2 \\ &\iff (1+x^2)(1+y^2)(1-xy)^2 \\ &\ge (1-x^2)(1-y^2)(1+xy)^2 \\ &\iff (1+x^2y^2+x^2+y^2)(1+x^2y^2-2xy) \end{aligned}$$
$$\geq (1 + x^2 y^2 - x^2 - y^2)(1 + x^2 y^2 + 2xy).$$
Let $z = 1 + x^2 y^2$, $t = x^2 + y^2$, $w = 2xy$, then
$$\begin{align*} (z+t)(z-w) &\ge (z-t)(z+w) \iff zt \ge zw \iff t \ge w \iff \\ &\Longleftrightarrow x^2+y^2 \ge 2xy \iff (x-y)^2 \ge 0, \end{align*}$$
which finishes the proof of the lemma.
By the lemma,
$$\left(\frac{1+m^2}{1-m^2} \cdot \frac{1+n^2}{1-n^2}\right) \left(\frac{1+n^2}{1-n^2} \cdot \frac{1+k^2}{1-k^2}\right) \left(\frac{1+k^2}{1-k^2} \cdot \frac{1+m^2}{1-m^2}\right) \ge \\ \ge \left(\frac{1+mn}{1-mn}\right)^2 \left(\frac{1+nk}{1-nk}\right)^2 \left(\frac{1+km}{1-km}\right)^2,$$
i.e.
$$\left( \frac{1+m^2}{1-m^2} \cdot \frac{1+n^2}{1-n^2} \cdot \frac{1+k^2}{1-k^2} \right)^2 \ge \left( \frac{1+mn}{1-mn} \cdot \frac{1+nk}{1-nk} \cdot \frac{1+km}{1-km} \right)^2,$$
with equality if and only if $m = n = k$, which gives the required statement of the problem.