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In this problem, you can actually demonstrate $P(k+1)$ without invoking $P(k)$! That means that you really haven't done an induction proof, but have rather done a direct proof: \[ (k+1)^2 \ge 2(k+1)+3 \] \[ \iff \] \[ k^2+2k+1 \ge 2k+5 \] \[ \iff \] \[ k^2 \ge 4 \]
So $P(k+1)$ for $k\ge 2$ means $P(n)$ $\forall n\ge 3$