Isothermal vs. adiabatic compression of gas in terms of required energy
To solve this, try to use what I call the "graphical apparatus". For an isothermal process:
$$
\begin{align}
PV&amp;=\text{constant}\\
P\mathrm{d}V&amp;=-V\mathrm{d}P\\
\frac{\mathrm{d}P}{\mathrm{d}V}&amp;=-\frac{P}{V}\\
\end{align}$$
for adiabatic process:
$$
\begin{align}
PV^\gamma&amp;=\text{constant}\\
\frac{\mathrm{d}P}{\mathrm{d}V}&amp;=-\gamma\frac PV
\end{align}$$
Therefore, starting at the same point on a P-V graph, the curves for an adiabatic and isothermal processes will diverge and the adiabatic curve will have a steeper slope. For the same reduction in volume (the graph in the picture is for expansion, not for contraction. In case of contraction, the curves will be reversed, i.e. adiabatic curve will be above the isothermal curve, and will enclose greater area under it for the same reduction in pressure), more area will be enclosed by adiabatic, and since the area $\int P\mathrm{d}V$ gives the work required, isothermal work is smaller than adiabatic for the same reduction in volume.  

Your argument is correct. To provide more mathematical support to it, you can observe the fact that it is both increase in temperature and reduction in volume which increases the pressure in adiabatic process and only reduction in volume increases pressure in isothermal process. The exponent of volume in adiabatic equation ($PV^\gamma=K$) is $\gamma&gt;1$, as compared to 1 in the isothermal equation. Hence, pressure change is more sensitive to volume change and is larger in magnitude.