How prove this geometry inequality in $\Delta ABC$,if $AD$ bisects $\angle BAC$, $MD$ bisects $\angle ADB$, $ND$ bisects $\angle ADC$,prove that
$$\dfrac{1}{BM}+\dfrac{1}{CN}\le\dfrac{4}{MN}$$

my idea:use if $AD$ bisects $\angle BAC$,then we have 
$$\dfrac{AB}{AC}=\dfrac{BD}{DC}$$
 A: Since you can scale the triangle without changing the ratio, it is convenient to put $D$ at the origin $(0,0)$ and $A$ at $(-1,0)$. Then your problem looks like this

I can assume $0<\alpha<\phi<\frac{\pi}{2}$.
Then the lines in the figure are given by the following simple formulas
$AC:\quad x\sin\alpha-y\cos\alpha+\sin\alpha=0$,
$AB:\quad x\sin\alpha+y\cos\alpha+\sin\alpha=0$,
$BC:\quad x\sin\phi-y\cos\phi=0$,
$DN:\quad \displaystyle x\sin\frac{\phi+\pi}{2}-y\cos\frac{\phi+\pi}{2}=0$,
$DM:\quad \displaystyle x\sin\frac{\phi}{2}-y\cos\frac{\phi}{2}=0$.
It is now easy to find the coordinates of the points by intersecting the lines, which gives:
$\displaystyle N=\left(\frac{\tan\alpha}{\tan(\phi+\pi)/2-\tan\alpha}, \frac{\tan\alpha\tan(\phi+\pi)/2}{\tan(\phi+\pi)/2-\tan\alpha}\right)$,
$M=\displaystyle\left(-\frac{\tan\alpha}{\tan\phi/2+\tan\alpha}, -\frac{\tan\alpha\tan\phi/2}{\tan\phi/2+\tan\alpha}\right)$,
$B=\displaystyle\left(-\frac{\tan\alpha}{\tan\phi+\tan\alpha}, -\frac{\tan\alpha\tan\phi}{\tan\phi+\tan\alpha}\right)$,
$C=\displaystyle\left(\frac{\tan\alpha}{\tan\phi-\tan\alpha}, \frac{\tan\alpha\tan\phi}{\tan\phi-\tan\alpha}\right)$,
Plugging these into the usual Euclidean distance formula gives $MN$, $BM$ and $CN$. After some simplification with mathematica, I get the following
$MN=\displaystyle\frac{2\sin\alpha\sqrt{1+\sin\phi\sin2\alpha}}{\sin\phi+\sin2\alpha}$,
$BM=\displaystyle\frac{\sin\alpha\sin\phi/2}{\sin(\phi+\alpha)\sin(\phi/2+\alpha)}$,
$CN=\displaystyle\frac{\sin\alpha\cos\phi/2}{\sin(\phi-\alpha)\cos(\phi/2-\alpha)}$.
Then $R(\alpha,\phi)=\displaystyle\frac{MN}{BM}+\frac{MN}{CN}$ is a function of two variables $\alpha,\phi$ and it looks like this

And here is another plot showing just the bit that sticks out above 4

It is obvious from these plots that the maximum of $R(\alpha,\phi)$ occurs on the diagonal $\alpha=\phi$, where 
$$r(\alpha)=R(\alpha,\alpha)= 2\sqrt{2}\cos\alpha \sqrt{2 + \cos\alpha - \cos3\alpha}.$$
It reaches the maximum at $\alpha_m$ which satisfies $-8 \sin\alpha_m - 2 \sin2\alpha_m + 5 \sin4\alpha_m=0$. This equation can be solved in mathematica analytically but the result is pretty ugly. For numerical values, I get $\alpha_m=0.4371582260685195$ and $r(\alpha_m)=4.170978697039997$. So, the correct inequality is
$$\frac{1}{BM}+\frac{1}{CN}\le\frac{r(\alpha_m)}{MN}$$
A: Hint:We consider $\Delta ABC$ as triangle in $\mathbb{R}^2$($O$ origin is midpoint of $BC$)and encircling it in the ellipse with the property that its two foci are $B(-a,0),C(a,0)$ and passes $A(x_0,y_0)$.
canonical equation of the ellipse is $\frac{x^2}{d^2}$+${y^2}\frac{1-\frac{x_0^2}{d^2}}{y_0^2}=1$,$d=\frac{1}{2}(\sqrt{(x_0-a)^2+y_0^2}+\sqrt{(x_0+a)^2+y_0^2})$, then the normal line to tangent line of the ellipse at $A$ is bisector of $\angle BAC$(here) now we find Coordinates of $D$,similarly we find $M$ and $N$,now compute $BM,CN,MN$
