we know that $\angle ABT + \angle ADT = \angle BCD=\angle BCA + \angle ACD.$ so we have $\angle ADT \ge \angle ACD$ or $\angle ABT \ge \angle ADT$. assume that the first one is true. consider the circle $\omega$ tangent to $AD$ at $D$ and passing through $T$. clearly this circle meets line $AC$ at point $M$ so that $AC \ge AM$ because $\angle ADT \ge \angle ACD$ and $C$ is not inside $\omega$ . now we have $AD^2={AT}\times{AM}$. so that $4AD^2=4{AT}\times{AM}\le(AT+AM)^2$ so $2AD \le AT+AM \le AT+AC$. and it's done.