How should I interpret this thermodynamic inequality? I'm a little bit confused over the following inequality: 
$$ dS > \frac{\delta Q_{irrev}}{T} $$
An infinitesimal change in entropy is defined in this way: $$  dS = \frac{\delta Q_{rev}}{T} $$
Such that $$   \frac{\delta Q_{rev}}{T} > \frac{\delta Q_{irrev}}{T} $$
And this would imply that $$ \delta Q_{rev} > \delta Q_{irrev} $$
I find this a little confusing, because I would argue that, for example, in order to raise a system from an initial temperature to a final temperature a certain, fixed amount of heat is needed, regardless whether the process is reversible or not. If the process is irreversible, heat is transfered less efficient, so more heat in total would be needed to supply the fixed amount to the system, thereby implying that: 
$$ \delta Q_{rev} < \delta Q_{irrev} $$
Can someone explain to me why my reasoning is wrong? 
 A: I think the confusion can be removed if we change the position of the subscripts:
Let us say we give a certain amount of heat $\delta Q$ to the system. If we do this via a reversible process, the resultant change in entropy of the system is given as $\delta S_{rev} = \frac{\delta Q}{T}$
Now, consider if we provide the same amount of heat to the system, but now via an irreversible process. Now, the resultant rise in entropy of the system is given as $\delta S_{irrev} >\frac{\delta Q}{T}$.
As we can see, for the same amount of heat $\delta Q$ provided to the system, the entropy rises more if we follow an irreversible process, than it does if we follow a reversible one: $\delta S_{irrev} > \delta S_{rev}$.
The second last equation you wrote is also correct : For the same rise in entropy of the system, we would need to provide less heat if we are going to provide it via an irreversible process.
A: My answer to this question differs substantially from the other answers that have been given, and is based on the developments presented in the following:  Thermodynamics by Enrico Fermi and Fundamentals of Engineering Thermodynamics by Moran et al.  
For an irreversible process between two thermodynamic equilibrium states of a closed system (no mass entering or leaving the system), the change in entropy between the two thermodynamic equilibrium end states satisfies the following inequality:  $$\Delta S>\int{\frac{dQ_{irrev}}{T_{boundary}}}$$where $dQ_{irrev}$ represents a differential amount of heat passing through the boundary between the surroundings and the system (say during some time interval dt during the irreversible process) and $T_{boundary}$ represents the temperature at the boundary through which $dQ_{irrev}$ is passing (during that same time interval).  Note that in an irreversible process, the temperature within the system typically varies with spatial position, so there is no one single average system temperature that one can use in the denominator of the integral.  Clausius, when he developed this inequality, determined that the correct temperature to use in his inequality is the system temperature at the boundary with the surroundings (through which the heat is passing).  However, this distinction has rarely been articulated, or even recognized, by most authors of thermodynamics texts.  This has caused countless students endless confusion over the past few centuries.
The change in entropy for the system between the same two thermodynamic equilibrium end states is:  $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$This equation applies to any reversible path between the two end states (there are an infinite number of reversible paths to select from) and T is the (spatially uniform) system temperature during the heat transfer of $dQ_{rev}$.  Note that, for a reversible path, the temperature at the boundary between the system and surroundings is equal to the system temperature T throughout the reversible path:  $$T_{boundary}=T$$Note also that $dQ_{rev}$ is not equal to $dQ_{irrev}$ along these paths, and that the corresponding boundary temperatures for the reversible and irreversible paths are likewise not equal.  The paths, heat flows, and boundary temperatures are totally different.
A: Your intuition is incorrect. See for instance the Joule expansion, in which a gas expands adiabatically after you open a compartment. Here $dQ_{irr}$ is zero. But to calculate the change in entropy you usually use an isothermal expansion in which all heat is transformed into work, see here
In this case, you can clearly see that $dQ_{rev}>dQ_{irr}=0$
