Pressure and LeChatelier's Principle
Le Chatelier's principle is not about pressure, it is about concentrations and temperature (Please also see the good article on Wikipedia for Le Chatelier's principle). 

Le Chatelier's principle is also called "The Equilibrium Law", that can be used to predict the effect of a change in conditions on a chemical equilibrium. Chemical equilibrium means that reaction is reversible: it can go as forward, so backwards. 

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

Lets imagine that $\rm A,\,B,\,C,\,D$ are gases. So you wrote correctly that a pressure increase will cause the reaction equilibrium to shift to the "right" - more products $\rm C,\,D$ will be formed. When we increase the pressure in the system, it means that we reduce volume. We squeeze our system, and there is less volume for molecules to be in. They hit each other more frequently, and the probability of product formation increases. When you reduce the volume by increasing the pressure, you basically increase the concentration of reagents. The same amount of atoms or molecules occupy lesser volume.


I'll make a silly comparison. Let's take this reaction:
$$\rm Boy+Girl\longleftrightarrow Pair$$
If there is too little space on the dance-floor to dance on your own it could be a good idea to find a partner to reduce the volume you both occupied to feel much more comfortable. But when the other people start to leave the club you can unpair with your partner...


Lets take 2 mol of $\rm A$, 3 mol of $\rm B$, 1 mol of $\rm C$, 1 mol of $\rm D$ at room temperature. We will put it in the jar of $\rm 156.8\,L$ ($\rm (2+3+1+1)\,mol \cdot 22.4\, L\cdot mol^{-1}=156.8\,L$). Pressure in the jar is $\rm 1\,atm$.


$P(A)=X(A)\cdot P(total)=2/7\cdot1=0.286\,atm$
$P(B)=X(B)\cdot P(total)=3/7\cdot 1=0.439\,atm$
$P(C)=X(C)\cdot P(total)=1/7\cdot 1=0.143\,atm$
$P(D)=X(D)\cdot P(total)=1/7\cdot 1=0.143\,atm$


Our system is in equilibrium, so we can calculate Equilibrium constant: $$K_P=\dfrac{P(C)^1P(D)^1}{P(A)^2P(B)^3}=\dfrac{0.143\cdot0.143}{0.286^2\cdot0.439^3}=3.17\,atm^{-3}$$
$$K_P=\dfrac{(X(C)\cdot P(total))^1(X(D)\cdot P(total))^1}{(X(A)\cdot P(total))^2(X(B)\cdot P(total))^3}=\dfrac{X(C)^1X(D)^1P^{\Delta n}}{X(A)^2X(B)^3};\,\Delta n=n(C)+n(D)-n(A)-n(B)=1+1-2-3=-3$$

Now lets double the pressure in the jar, by reducing the volume, the temperature did not change so $K_P$ stays the same. Lets calculate the molar concentrations from:

$$K_P=3.17\,atm^{-3}=\dfrac{X'(C)^1X'(D)^1\cdot (2\,atm)^{-3}}{X'(A)^2X'(B)^3},$$
 where 


$\upsilon'(A)=2\,mol-2x\,mol$
$\upsilon'(B)=3\,mol-3x\,mol$
$\upsilon'(C)=1\,mol+x\,mol$
$\upsilon'(D)=1\,mol+x\,mol$


Total mols: $\sum=(1+x)+(1+x)+(2-2x)+(3-3x)=(7-3x)\,mol$


$X'(A)=(2-2x)/(7-3x)$
$X'(B)=(3-3x)/(7-3x)$
$X'(C)=(1+x)/(7-3x)$
$X'(D)=(1+x)/(7-3x)$


Solving the eqution gives $x=0.325\,mol$, so now:


$\upsilon'(A)=1.35\,mol$
$\upsilon'(B)=2.03\,mol$
$\upsilon'(C)=1.33\,mol$
$\upsilon'(D)=1.33\,mol$


But it the begining we has $A$=2 mol, $B$=3 mol, $C$=1 mol, $D$=1 mol. The reactions shifted to the right. 


$P'(A)=0.45\,atm$
$P'(B)=0.67\,atm$
$P'(C)=0.44\,atm$
$P'(D)=0.44\,atm$




Ok, but when we add inert gas, what happens:

$$\rm 2A+3B+inert\,gas \longleftrightarrow C+D+inert\,gas $$

It does not do anything because even though the pressure increased, the probabilty for 2 atoms $\rm A$ to hit 3 atoms $\rm B$ does not change. The partial pressures stay the same. Why is it so? It can be easily shown. For example we have this reaction:
$$\rm A_g+B_g\longleftrightarrow C_g$$

Lets say we put this reaction in a jar of 134.4 L volume at room temperature, we have 3 mol of $\rm A$, 2 mol $\rm B$, and one 1 mol $\rm C$. 6 mol of gases being put in 134.4 L at RT will give us 1 atm pressure. 
$$6\,{\rm mol} \cdot 22.4\,{\rm L\cdot mol^{-1}}=134.4\,{\rm L}$$

The partial pressures are: 


$ P(A)=3/6\cdot P_{total}=0.5\cdot1\,atm$
$ P(B)=2/6\cdot P_{total}=0.33\cdot1\,atm$
$P(C)=1/6\cdot P_{total}=0.17\cdot1\,atm$ 


Now we will add 1 mol of inert gas:
$$\rm A_g+B_g+inert\,gas\longleftrightarrow C_g+inert\,gas$$

Now, the partial pressures are (pressure in the jar increased, because quantatiy of molecules increased, but volume stays the same): 


$P(A)=3/7\cdot P_{total}=3/7\cdot 1.17\,atm=0.5\,atm$
$P(B)=2/7\cdot P_{total}=2/7\cdot 1.17\,atm=0.33\,atm$
$P(C)=1/7\cdot P_{total}=1/7\cdot 1.17\,atm=0.17\,atm$


As you can see, the partial pressures did not change (concentration did not change). That is why there is no effect.


Let's take this reaction:
$$\rm Boy+Girl+Cat\longleftrightarrow Pair+Cat$$
You cannot dance with a cat. Adding extra cats on the dance-floor will increase the pressure in the crowd, but will not increase the probability of formation of a dancing pair.
 



Let's imagine that $\rm A$ and $\rm B$ are liquids, and see what a pressure increase will cause. Usually it does not cause a lot, because it is very difficult to squeeze liquids. No matter how you try, the volume which molecules of liquid occupy stays the same. So increasing the pressure in reaction:

$$\rm A_\ell+B_\ell\longleftrightarrow C_\ell$$

will not do practically anything, because concentrations of reagents will not increase under the pressure.

When I wrote $\Delta H=-100\,kJ\cdot mol^{-1}$ (the enthalpy is negative), it means that in reaction the heat is released.  

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

So if we decrease temperature in the jar, reaction will move to the right. It can be understood using this primitive logic. Assume that heat is some substance, that could be treated as "product". If you remove this so called "product" (heat, by temperature reduction) you constantly force reactin to happen from left to right. In reality the equilibrium is just shifted from left to right.

If we increase temperature, it is like we added more of this so called "product" (heat) in the jar, so reaction is moved to the left. 

If $\Delta H&gt;0$, you need to think about heat not as "product", but as "reagent".

For better understanding try to "dive" into Equilibrium thermodynamics.