Consider the following:
ω = 1/√(LC)
Xc = 1/ωC = √(L/C)
Since the circuit is at resonance, or almost at resonance, the current in the tank is constant regardless of L and C.
Let the current be I. Then,
Vc = I×√(L/C)
Now for a given capacitance, if the resonance is lowered by using a larger inductor, it leads to larger voltage across the capacitor. Since larger voltage rating capacitors do not have large current capability(as mentioned in the text also), we do not want to lower the resonance unnecessarily. Lower resonance is associated with larger circulating energy in the tank.