Consider the slopes ($dP/dV$) for both processes on a P-V diagram:

1. Isothermal: $PV=C⟹P=C/V$. Slope $=-P/V$.
2. Adiabatic: $P{V}^{\gamma }=C⟹P=C/V^{\gamma }$. Slope $=-\gamma (P/V)$. Since $\gamma >1$ (it is $5/3$ for monoatomic, $7/5$ for diatomic), the magnitude of the adiabatic slope is always greater than the isothermal slope. This means the adiabatic curve is steeper.