Belorusija 2012

Let $a=BC$, $b=AC$, $c=AB$, and $\angle A=\alpha$, $\angle B=\beta$, $\angle C=\gamma$. Let $A{H}_1$, $B{H}_2$, $C{H}_3$ be altitudes, $A{L}_1$, $B{L}_2$, $C{L}_3$ be angle bisectors, and $A{M}_1$, $B{M}_2$, $C{M}_3$ be medians of $△ABC$. Then $$\frac{AI}{L_{1}I}=\frac{b+c}{a},\frac{BI}{L_{2}I}=\frac{a+c}{b},\frac{CI}{L_{3}I}=\frac{a+b}{c}.$$ Since $AG:G{M}_1=BG:G{M}_2=CG:G{M}_3=2:1$, we have $$f(G)=\frac{AG}{M_{1}G}\cdot \frac{BG}{M_{2}G}\cdot \frac{CG}{M_{3}G}=2\cdot 2\cdot 2=8.$$ Hence, $$\begin{array}{rl}f(I)& =\frac{AI}{L_{1}I}\cdot \frac{BI}{L_{2}I}\cdot \frac{CI}{L_{3}I}=\frac{b+c}{a}\cdot \frac{a+c}{b}\cdot \frac{a+b}{c}\ge \frac{2\sqrt{bc}}{a}\cdot \frac{2\sqrt{ac}}{b}\cdot \frac{2\sqrt{ab}}{c}=8=f(G),\end{array}$$ i.e. $f(I)\ge f(G)$ with the equality if and only if $a=b=c$.

Further,

$$\begin{array}{rl}H{H}_1& =B{H}_1\cdot \mathrm{tg}\angle HB{H}_1=\\ & =(AB\cos \beta )\cdot \mathrm{tg}(90^{\circ }-\gamma )=\frac{c\cdot \cos \beta \cdot \cos \gamma }{\sin \gamma }\end{array}$$ and $$AH=\frac{A{H}_2}{\cos \angle HA{H}_2}=\frac{AB\cdot \cos \alpha }{\cos (90^{\circ }-\gamma )}=\frac{c\cdot \cos \alpha }{\sin \gamma }.$$ So we get $\frac{AH}{H_{1}H}=\frac{\cos \alpha }{\cos \beta \cos \gamma }$. Similarly, $\frac{BH}{H_{2}H}=\frac{\cos \beta }{\cos \alpha \cos \gamma }$ and $\frac{CH}{H_{3}H}=\frac{\cos \gamma }{\cos \alpha \cos \beta }$. $$\text{Therefore, }f(H)=\frac{AH}{H_{1}H}\cdot \frac{BH}{H_{2}H}\cdot \frac{CH}{H_{3}H}=\frac{1}{\cos \alpha \cos \beta \cos \gamma }.$$ Let $m=\mathrm{tg}\frac{\alpha }{2}$, $n=\mathrm{tg}\frac{\beta }{2}$ and $k=\mathrm{tg}\frac{\gamma }{2}$. 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 $$\begin{array}{rl}\frac{b+c}{a}& =\frac{b}{a}+\frac{c}{a}=\frac{\sin \beta }{\sin \alpha }+\frac{\sin \gamma }{\sin \alpha }=\frac{2\sin \frac{\beta +\gamma }{2}\cos \frac{\beta -\gamma }{2}}{\sin (180^{\circ }-(\beta +\gamma ))}=\\ & =\frac{2\sin \frac{\beta +\gamma }{2}\cos \frac{\beta -\gamma }{2}}{\sin (\beta +\gamma )}=\frac{\cos \frac{\beta -\gamma }{2}}{\cos \frac{\beta +\gamma }{2}}=\frac{\cos \frac{\beta }{2}\cos \frac{\gamma }{2}+\sin \frac{\beta }{2}\sin \frac{\gamma }{2}}{\cos \frac{\beta }{2}\cos \frac{\gamma }{2}-\sin \frac{\beta }{2}\sin \frac{\gamma }{2}}=\\ & =\frac{1+\mathrm{tg}\frac{\beta }{2}\mathrm{tg}\frac{\gamma }{2}}{1-\mathrm{tg}\frac{\beta }{2}\mathrm{tg}\frac{\gamma }{2}}=\frac{1+nk}{1-nk}.\end{array}$$ 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, $$f(H)\ge f(I)⟺\frac{1+m^2}{1-m^2}\cdot \frac{1+n^2}{1-n^2}\cdot \frac{1+k^2}{1-k^2}\ge \frac{1+mn}{1-mn}\cdot \frac{1+nk}{1-nk}\cdot \frac{1+km}{1-km}$$ 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 $$\begin{array}{rl}\frac{1+x^2}{1-x^2}\cdot \frac{1+y^2}{1-y^2}& \ge {\left(\frac{1+xy}{1-xy}\right)}^2\\ & ⟺(1+x^2)(1+y^2)(1-xy{)}^2\\ & \ge (1-x^2)(1-y^2)(1+xy{)}^2\\ & ⟺(1+x^2{y}^2+x^2+y^2)(1+x^2{y}^2-2xy)\end{array}$$ $$\ge (1+x^2{y}^2-x^2-y^2)(1+x^2{y}^2+2xy).$$ Let $z=1+x^2{y}^2$, $t=x^2+y^2$, $w=2xy$, then $$\begin{array}{rl}(z+t)(z-w)& \ge (z-t)(z+w)⟺zt\ge zw⟺t\ge w⟺\\ & ⟺x^2+y^2\ge 2xy⟺(x-y{)}^2\ge 0,\end{array}$$ which finishes the proof of the lemma. By the lemma, $$\begin{gathered} \left(\frac{1+m^2}{1-m^2}\cdot \frac{1+n^2}{1-n^2}\right)\left(\frac{1+n^2}{1-n^2}\cdot \frac{1+k^2}{1-k^2}\right)\left(\frac{1+k^2}{1-k^2}\cdot \frac{1+m^2}{1-m^2}\right)\ge \\ \ge {\left(\frac{1+mn}{1-mn}\right)}^2{\left(\frac{1+nk}{1-nk}\right)}^2{\left(\frac{1+km}{1-km}\right)}^2, \end{gathered}$$ 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.