How prove this geometry inequality

Question

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 $$\dfrac{AB}{AC}=\dfrac{BD}{DC}$$

Answer 1

Your statement cannot be proven because it is incorrect. Take the following coordinates:

\begin{align*} A &= \begin{pmatrix}0\\1\end{pmatrix} & B &= \begin{pmatrix}1\\0\end{pmatrix} & C &= \begin{pmatrix}\lambda\\0\end{pmatrix} \end{align*}

Then in the limit $\lambda\to+\infty$ you get

$$\lim_{\lambda\to+\infty} \frac{MN}{BM} + \frac{MN}{CN} = \lim_{\lambda\to+\infty}\frac{MN}{BM} = \sqrt{\sqrt{4\sqrt2 + 8} + 4\sqrt2 + 8} \approx 4.1656 > 4 $$

This result can either be obtained in a CAS of your choice, or – if you don't need the exact number and trust your floating point computations – in some suitable geometry program. In either case, the bottom line is that if you just move $C$ far enough, and have $\angle CBA$ sufficiently large, then the equation you stated will be violated.

Note that this is not the maximal violation. The nearest local maximum to this situation appears to be for $A\approx(0.16371,1)$, where the ratio reaches a value of $\frac{MN}{BM}\approx4.17098$. This was found using numeric optimization, so a nice symbolic expression for this is still missing. Chances are that you might be able to prove

$$\frac1{BM}+\frac1{CN}<\frac{4.2}{MN}$$

or something along these lines. To do this, you'd have to show that this local maximum is also the global one. The underlying function, in terms of the $x$ coordinate of both $A$ and $C$, is

sqrt(4*((2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx^2/(4*(2*Ax - Cx)^2*Ax*Cx^2 -
(2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax -
Cx)^2) - 2)*Cx + sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx) -
(2*(2*Ax - Cx)^2*Cx^3 + (2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx^2)/(4*(Ax -
Cx)*(2*Ax - Cx)^2*Cx^2 - (2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 +
4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) -
2)^2*Cx^2))*Cx))^2 + (2*(2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Ax*Cx^2/(4*(2*Ax -
Cx)^2*Ax*Cx^2 - (2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax -
1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx + sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax
- Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) -
2)^2*Cx^2))*Cx) - (2*(Ax - Cx)*(2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax -
Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx^2 +
(2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax -
Cx)^2) - 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) -
2)^2*Cx^2))*Cx^2)/(4*(Ax - Cx)*(2*Ax - Cx)^2*Cx^2 - (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx))^2)*(1/sqrt((Cx +
(2*(Ax - Cx)*(2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx^2 + (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx^2)/(4*(Ax - Cx)*(2*Ax
- Cx)^2*Cx^2 - (2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax -
1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax
- Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) -
2)^2*Cx^2))*Cx))^2 + 4*(2*(2*Ax - Cx)^2*Cx^3 + (2*Ax - Cx)*(2*(Ax -
Cx)*Ax - 2*(2*Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax -
Cx)^2) - 2)*Cx^2)^2/(4*(Ax - Cx)*(2*Ax - Cx)^2*Cx^2 - (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx)^2) + 1/sqrt(4*(2*Ax -
Cx)^2*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 +
4*(2*Ax - Cx)^2) - 2)^2*Ax^2*Cx^4/(4*(2*Ax - Cx)^2*Ax*Cx^2 - (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx + sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx)^2 + 4*(2*Ax -
Cx)^2*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 +
4*(2*Ax - Cx)^2) - 2)^2*Cx^4/(4*(2*Ax - Cx)^2*Ax*Cx^2 - (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx + sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx)^2))

So “all you have to do” is compute the partial derivatives of this beast, solve for them being zero at the same time, enumerate all critical points, and compare them numerically. Good luck finding roots for the partial derivatives, though!

Answer 2

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

$$\frac{1}{BM}+\frac{1}{CN}\le\frac{r(\alpha_m)}{MN}$$

Answer 3

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 $BM,CN,MN$