You don't need to forward bias the B-E junction. Whether you do depends on what you want the transistor to do. To keep it off, you want to not forward bias the B-E junction. As for the C-B junction, keeping that reverse biased is fundamental to how BJTs operate.
A BJT is basically a reverse biased junction that can be made selectively leaky. You apply a voltage accross C-E and with the base open nothing happens. The reverse biased junction doesn't allow any (except for small leakage we will ignore) current to flow. However, the special property of a BJT is that a little current thru the base messes up the insulating capability of the reverse biased junction. The gain, and hence the useful properties, of a BJT come from the fact that is only takes a little current to muck up the reverse biased junction such that it allows a lot more C-E current to flow. This ratio of C-E current to B-E current is the basic gain of the transistor. It can be as low as 5-10 in big mongo power transistors and 100s in high gain signal transistors.
The B-E junction also looks like a diode to the external circuit. It will have a forward voltage drop when conducting just like a regular diode. In silicon, this is 500-750 mV for most non-extreme applications.
If you want to use the transistor as a switch (either as full off or full on as you can make it) then you have to make sure there is no base current in the off case, and plenty enough to support the desired collector current in the on case. Driving the base to the emitter voltage is a good way to make sure the transistor is off. To turn it fully on, you need to provide at least 1/gain of the desired collector current.
In other cases, a BJT might be used in "linear" (it's often rather non-linear, but this is the term used to mean in-between mode or not-switch mode) mode, like a audio amplifier. In that case you want to always keep it somewhat on and have the input signal change its operating point. If done right, this can amplify the signal. Different configurations give you voltage gain, or current gain, or some combination. In these cases, biasing the transistor refers to keeping it somewhere in the middle of the operating range so that a little input signal can change the output both ways. Biasing is basically setting up the DC operating point.
Remember the V(CE) vs I(C) curve and consider an NPN transistor. Thus forward biasing simply means there is positive voltage on the p-type junction relative to an n-type junction.
Thus, in our case forward biasing the BC junction means having a voltage V(BC) > 0.
Now, when collector voltage V(C) drops below the base voltage V(B) and forward biases the collector‐base junction (when V(BC) > 0):
The strong electric field which opposes the movement of the charged particles into the collector is now weakened.
There is now a numerous charged particles into the collector, hence current increases and the current gain factor, β,
decreases.
This corresponds to a logical more of "0" where the transistor acts as a switch, showing a low resistance path for current conduction.
Best Answer
The easiest way to drive a transistor into saturation is to ensure you are providing copious amounts of base current, say \$I_C/10\$.
You can also schottky clamp the base-collector terminal so that you only drive the transistor into being on the edge of saturation, and thus having a faster turn-off time.