= Solution
The positive-part function is <convex>, so pathwise <Jensen inequality> gives
$$
\left(\frac1T\sum_{t=1}^T S_t-K\right)^+
\leq\frac1T\sum_{t=1}^T(S_t-K)^+.
$$
To compare the <Asian option> price with <European call option> prices at different dates, carry each earlier payoff forward using the <numéraire>. Purchase $1/T$ of each <replicating strategy> for maturity $t$, and, when its payoff is received, reinvest it in $N$ until $T$. This is a <self-financing portfolio>, with terminal wealth
$$
\frac1T\sum_{t=1}^T (S_t-K)^+\frac{N_T}{N_t}
\geq\frac1T\sum_{t=1}^T(S_t-K)^+
\geq\left(\frac1T\sum_{t=1}^T S_t-K\right)^+.
$$
The first inequality uses $N_T\geq N_t$ and nonnegative payoffs. Since the <Asian option> is replicable in the <complete market>, absence of <arbitrage> makes its replication cost no larger than this <superhedging> cost. \b[Therefore]
$$
\boxed{A(T,K)\leq\frac1T\sum_{t=1}^T C(t,K).}
$$
Equivalently, divide the first pathwise bound by $N_T$, use $(S_t-K)^+/N_T\leq(S_t-K)^+/N_t$, and take <expectations> under the <numéraire> <equivalent martingale measure>. Merely averaging earlier payoffs without reinvestment would miss the difference in payment dates.
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