Prove there exists a point O such that $ OB \leq \frac {2}{\sqrt {3}} BY, OD \leq \frac {2}{\sqrt {3}} DZ, OF \leq \frac {2}{\sqrt {3}} FX$ 
Let the incircle of BDF touch DF, FB, BD at X, Y, Z respectively. Prove  there exists a point O such that $$ OB \leq \frac {2}{\sqrt {3}} BY, OD \leq \frac {2}{\sqrt {3}} DZ,  OF \leq \frac {2}{\sqrt {3}} FX$$.

It seem interesting,How prove it?
 A: I'm a big fan of the tangent half-angle formula. Without loss of generality, the incircle is the unit circle, rotated such that none of $X,Y,Z$ lies at $(-1,0)$. Then one may choose coordinates as
\begin{align*}
X&=\frac1{1+x^2}\begin{pmatrix}1-x^2\\2x\end{pmatrix}&
Y&=\frac1{1+y^2}\begin{pmatrix}1-y^2\\2y\end{pmatrix}&
Z&=\frac1{1+z^2}\begin{pmatrix}1-z^2\\2z\end{pmatrix}\\
B&=\frac1{1+yz}\begin{pmatrix}1-yz\\y+z\end{pmatrix}&
D&=\frac1{1+xz}\begin{pmatrix}1-xz\\x+z\end{pmatrix}&
F&=\frac1{1+xy}\begin{pmatrix}1-xy\\x+y\end{pmatrix}
\end{align*}
I'll assume that the point $O$ you're after is the Gergonne point, where $BX$, $DY$ and $FZ$ intersect. Its coordinates are
$$O=\frac1{x^2y^2-x^2yz-xy^2z+x^2z^2-xyz^2+y^2z^2+x^2-xy+y^2-xz-yz+z^2}
\\[4ex]
\cdot\begin{pmatrix}
-x^2y^2+x^2yz+xy^2z-x^2z^2+xyz^2-y^2z^2+x^2-xy+y^2-xz-yz+z^2\\
x^2y+xy^2+x^2z-6xyz+y^2z+xz^2+yz^2
\end{pmatrix}$$
From this you can compute
\begin{align*}
\lVert OB\rVert^2 &= \scriptsize\frac{(z-y)^{4}\cdot(x^2+1)\cdot(x^2y^2+2x^2yz-4xy^2z+x^2z^2-4xyz^2+4y^2z^2+4x^2-4xy+y^2-4xz+2yz+z^2)}{(yz+1)^2\cdot(x^2y^2-x^2yz-xy^2z+x^2z^2-xyz^2+y^2z^2+x^2-xy+y^2-xz-yz+z^2)^2} \\
\lVert BY\rVert^2 &= \frac{(y-z)^2}{(yz+1)^2}
\end{align*}
Now you can translate your first inequality into
\begin{align*}
0 &\le 4\lVert BY\rVert^2 - 3\,\lVert OB\rVert^2 \\&=
\scriptsize\frac{(-y+z)^2\cdot(-x^2y^2+4x^2yz-2xy^2z-x^2z^2-2xyz^2+2y^2z^2+2x^2-2xy-y^2-2xz+4yz-z^2)^2}{(yz+1)^2\cdot(x^2y^2-x^2yz-xy^2z+x^2z^2-xyz^2+y^2z^2+x^2-xy+y^2-xz-yz+z^2)^2}\\&=
\scriptsize\left(\frac{(-y+z)\cdot(-x^2y^2+4x^2yz-2xy^2z-x^2z^2-2xyz^2+2y^2z^2+2x^2-2xy-y^2-2xz+4yz-z^2)}{(yz+1)\cdot(x^2y^2-x^2yz-xy^2z+x^2z^2-xyz^2+y^2z^2+x^2-xy+y^2-xz-yz+z^2)}\right)^2
\end{align*}
So that difference is a square of a real number, and therefore non-negative, proving your inequality. The same argument holds for the other two inequalities, simply by a change in variable names. Obviously this proof relies a lot on polynomial computations, and can benefit massively from the aid of computer algebra systems.
One could assume $y=0$, still without loss of generality, but the fact that generality is preserved is somewhat harder to see in this case, and the symmetry of the setup is lost. So use that if you have to do the computations by hand, but the above if you can trust a computer to do the tedious work for you.
