A_{1}

B_{1}

C_{1}

BC

CA

AB

ABC

\frac{A_{1}B}{A_{1}C}=\frac{B_{1}C}{B_{1}A}=\frac{C_{1}A}{C_{1}B}=k.

\left(\frac{AA_{1}}{BC}\right)^{2}+\left(\frac{BB_{1}}{CA}\right)^{2}+\left(\frac{CC_{1}}{AB}\right)^{2}\geqslant\left(\frac{3k}{k^{2}+1}\right)\cdot\left(\frac{2r}{R}\right)^{4},

R

r

ABC

BC=a

CA=b

AB=c

\frac{A_{1}B}{BC}=\frac{k}{k+1},~\frac{A_{1}C}{BC}=\frac{1}{k+1}.

AC^{2}\cdot A_{1}B+AB^{2}\cdot A_{1}C-AA_{1}^{2}\cdot BC=BC\cdot A_{1}B\cdot A_{1}C~\Leftrightarrow

\Leftrightarrow~b^{2}a\cdot\frac{k}{k+1}+c^{2}a\cdot\frac{1}{k+1}-AA_{1}^{2}\cdot a=a\cdot a\cdot\frac{k}{k+1}\cdot a\cdot\frac{1}{k+1}~\Leftrightarrow

\Leftrightarrow~\frac{b^{2}}{a^{2}}\cdot\frac{k}{k+1}+\frac{c^{2}}{a^{2}}\cdot\frac{1}{k+1}-\frac{AA_{1}^{2}}{a^{2}}=\frac{k}{(k+1)^{2}}~\Leftrightarrow

\Leftrightarrow~\left(\frac{AA_{1}}{BC}\right)^{2}=\left(\frac{AA_{1}}{a}\right)^{2}=\frac{b^{2}}{a^{2}}\cdot\frac{k}{k+1}+\frac{c^{2}}{a^{2}}\cdot\frac{1}{k+1}-\frac{k}{(k+1)^{2}}.

\left(\frac{BB_{1}}{AC}\right)^{2}=\frac{c^{2}}{b^{2}}\cdot\frac{k}{k+1}+\frac{a^{2}}{b^{2}}\cdot\frac{1}{k+1}-\frac{k}{(k+1)^{2}},

\left(\frac{CC_{1}}{AB}\right)^{2}=\frac{b^{2}}{c^{2}}\cdot\frac{k}{k+1}+\frac{a^{2}}{c^{2}}\cdot\frac{1}{k+1}-\frac{k}{(k+1)^{2}}.

\left(\frac{AA_{1}}{BC}\right)^{2}+\left(\frac{BB_{1}}{CA}\right)^{2}+\left(\frac{CC_{1}}{AB}\right)^{2}=

=\frac{k}{k+1}\left(\frac{b^{2}}{c^{2}}+\frac{c^{2}}{b^{2}}+\frac{a^{2}}{c^{2}}\right)+\frac{1}{k+1}\left(\frac{c^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}\right)-\frac{3k}{(k+1)^{2}}\geqslant

\geqslant\frac{k}{k+1}\cdot3\sqrt[3]{\frac{b^{2}c^{2}a^{2}}{a^{2}b^{2}c^{2}}}+\frac{1}{k+1}\cdot3\sqrt[3]{\frac{c^{2}a^{2}b^{2}}{a^{2}b^{2}c^{2}}}-\frac{3k}{(k+1)^{2}}=

=\frac{k}{k+1}\cdot3+\frac{1}{k+1}\cdot3-\frac{3k}{(k+1)^{2}}=\frac{3k}{k+1}+\frac{3}{k+1}-\frac{3k}{(k+1)^{2}}=3-\frac{3k}{(k+1)^{2}}

3-\frac{3k}{(k+1)^{2}}\geqslant\left(\frac{3k}{k^{2}+1}\right)^{2},

\frac{3k^{2}}{(k^{2}+1)^{2}}+\frac{k}{(k+1)^{2}}\leqslant1.

k^{2}+1\geqslant2k~\Rightarrow~(k^{2}+1)^{2}\geqslant4k^{2}~\Rightarrow~\frac{3k^{2}}{(k^{2}+1)^{2}}\leqslant\frac{3}{4},

k+1\geqslant2\sqrt{k}~\Rightarrow~(k+1)^{2}\geqslant4k~\Rightarrow~\frac{k}{(k+1)^{2}}\leqslant\frac{1}{4}.

\frac{3k^{2}}{(k^{2}+1)^{2}}+\frac{k}{(k+1)^{2}}\leqslant\frac{3}{4}+\frac{1}{4}=1.

\left(\frac{AA_{1}}{BC}\right)^{2}+\left(\frac{BB_{1}}{CA}\right)^{2}+\left(\frac{CC_{1}}{AB}\right)^{2}\geqslant3-\frac{3k}{(k+1)^{2}}\geqslant\left(\frac{3k}{k^{2}+1}\right)^{2}.

R\geqslant2r

1\geqslant\left(\frac{2r}{R}\right)^{4}

\left(\frac{3k}{k^{2}+1}\right)^{2}=\left(\frac{3k}{k^{2}+1}\right)^{2}\cdot1\geqslant\left(\frac{3k}{k^{2}+1}\right)^{2}\cdot\left(\frac{2r}{R}\right)^{4}.