An inequality on Cevians Let $\displaystyle AD$, $\displaystyle BE$, $\displaystyle CF$ be three cevians concurrent at $\displaystyle P$ inside the $\displaystyle \Delta ABC$. 
Prove or disprove that:
$$\displaystyle \dfrac{AD}{AP} + \dfrac{BE}{BP} + \dfrac{CF}{CP} \ge \dfrac{9}{2}$$
 A: The inequality is true!

It can be shown that (see proof at the end of the answer)
$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = 1$$
Note that this implies that
$$\displaystyle \frac{AP}{AD} + \frac{BP}{BE} + \frac{CP}{CF}  = 2$$
as $\displaystyle 1 - \frac{PD}{AD} = \frac{AP}{AD}$ etc.
Now we have the inequality (easily shown using $\text{AM} \ge \text{GM}$) that
$$\displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9$$
This shows that 
$$\displaystyle (\frac{AP}{AD} + \frac{BP}{BE} + \frac{CP}{CF})(\frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP}) \ge 9$$
and so
$$\displaystyle 2(\frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP}) \ge 9$$
i.e.
$$\displaystyle \frac{AD}{AP} + \frac{BE}{BP} + \frac{CF}{CP} \ge \frac{9}{2}$$
Note that the equality occurs only when $\displaystyle \frac{AP}{AD} = \frac{BP}{BE} = \frac{CP}{CF} = \frac{2}{3}$, which implies that $\displaystyle P$ is the centroid.

Proof
Let us try showing that
$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = 1$$
Consider the figure (repeated from above for convenience).

Note, if you are worried about acute triangle vs obtuse etc, a simple affine transformation will do to transform the triangle into an equilateral triangle.
Let X be the foot of perpendicular from A to BC and Y be the foot of the perpendicular from P to BC.
$\displaystyle \triangle AXD$ and $\triangle PYD$ are similar and thus
$\displaystyle \frac{PY}{AX} = \frac{PD}{AD}$.
Now $\displaystyle \frac{PY}{AX} = \frac{|\triangle PBC|}{|\triangle ABC|}$
where $\displaystyle |\triangle MNO|$ is the area of $\displaystyle \triangle MNO$.
Thus $\displaystyle \frac{PD}{AD} = \frac{|\triangle PBC|}{|\triangle ABC|}$
Similarly
$\displaystyle \frac{PE}{BE} = \frac{|\triangle PAC|}{|\triangle ABC|}$
$\displaystyle \frac{PF}{CF} = \frac{|\triangle PAB|}{|\triangle ABC|}$
Adding gives us
$$\displaystyle \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} = \frac{|\triangle PAB| + |\triangle PAC| + |\triangle PBC|}{|\triangle ABC|} = \frac{|\triangle ABC|}{|\triangle ABC|} = 1$$
