With respect to the previous poster, I'm not sure I follow your reasoning and it seems that you, too, may have some misunderstanding about what a transition state is.
Needaask, we've been down this road before. The definition of the transition state precludes the scenario you are describing. A transition state is a saddle point on the N-dimensional potential energy surface. Mathematically, along the reaction coordinate (which is the lowest energy path between the global minima that correspond to reactants and products) it is a local maximum value of energy (the derivative is equal to zero). This is the definition of a transition state. Therefore by definition it is not possible for the energy at any infinitesimal value to the left or right of the transition state along the reaction coordinate to have a higher energetic value. If it did, it would not be a saddle point on the PES, and therefore it would not qualify as a transition state. Note that it is possible for there to be other maxima along the reaction coordinate, in the case of a multistep mechanism that features several transition states and quasi-stable intermediates, but the energy of the activated complex still has to decrease before it may increase again to form a higher-energy transition state.
Important: the difference in energy between the reactants and the transition state is NOT necessarily (and not usually) equal to the energy required to break bonds necessary for the reaction to occur. This is because the activated complex at the transition state is characterized as a structure with both a breaking bond and a forming bond, simultaneously. If you were to undergo a process with complete breaking of a bond, followed by complete forming of a new bond, the energy required would typically be (far) greater than that characterized by the true transition state, where forming and breaking occur simultaneously. The former process would not be the lowest energy path, and so MOST reaction events would not occur via this mechanism. Although, strictly speaking, the probability of it happening is not zero. The reaction coordinate represents the most probable pathway that a reaction pathway can take. There are an infinite number of true pathways available, and the probability of them occuring depends on temperature and their relative positions on the PES. (In a real chemical reaction, not all pathways lead to the intended products, either. The PES is an idealization.) Do note, however, that the potential energy is a state function. The pathway taken only impacts kinetics. The total energy gained and released by all reaction pathways is the same, as long as the starting and ending points are identical.
Example: Suppose you have a reaction A-B + C --> A + B-C
Now we focus on two possible mechanisms: (1) A-B + C --> A + B + C --> A + B-C and (2) A-B + C --> A--B--C --> A + B-C
In (1), the bond between A-B is completely broken, yielding two very high energy (presumably) isolated atoms that are an infinite distance apart. Then a new bond between B-C forms. This process requires (all things being equal, and assuming only population of ground vibrational states) an input activation energy equal to the bond energy of A-B. This energy is equal to the difference in potential energy between A/B separated by their equilibrium bond distance, and A/B infinitely far apart, which can be determined roughly by any anharmonic oscillator approximation of your choice (Morse potential, say). Conversely, when B-C forms, a quantity of energy is released equal to the inverse process - bringing B/C infinitely far apart to their equilibrium bond distance. The difference in bond energies between A-B and B-C would be roughly equivalent to the overall change in energy from reactants to products, which would determine the overall thermodynamic spontaneity of the reaction (endergonic or exergonic, depending on the sign). Of course, here I am treating enthalpy only and ignoring entropy, but for this simple hypothetical reaction, it's all right I think to paint a conceptual picture. Note also that I'm neglecting any energy of having A or C sitting around all by themselves. In a low pressure gas phase reaction, this is probably fine to do for sake of argument, but in solution there will be solvent interactions (enthalpies) to worry about that could not be neglected Note finally that for this process to happen SOME collision must occur because the energy has to come from somewhere; this could be some spectator gas molecule.
In (2), a different process happens: the bond between A-B begins to break, but at the same time, B-C begins to form. The transition state is a point somewhere in the middle, where A-B is "half" broken and B-C is "half" formed. (It won't really probably be "half" and "half", but whatever.) Note this is not the same as a hypothetical molecule A-B-C! It should be readily apparent that the energy of this transition state A--B--C MUST be equal to or lower than the energy of A + B + C, all isolated. To understand why, just take A-B for a moment. Now start to pull B away. Gradually the potential energy increases until at large separations some asymptotic limit is reached. This asymptotic limit is the maximum energy of the system attainable by pulling B away from A. Likewise, start with B-C and pull B away. Same thing, same energy. So at the most, A--B--C has the energy of A + B + C in the limit of the "--" being infinitely long. Shorter, and there has to be some stabilization, because bonds stabilize. Of course, if you were to shrink both "--" such that they were on the order of A-B and B-C (A-B-C), you'd probably have a complex with higher energy than A + B + C, but this isn't the transition state. This is all to say, that A--B--C must be lower energy than A + B + C, else no reaction would proceed through A--B--C. It wouldn't be along the most probably reaction coordinate, because reactions proceed through the path of least resistance.
Anyway, what you end up with is a transition state energy that is more than either the reactant or the product but less than blasting everything to individual atoms and then reassembling them. The latter process, by the way, is effectively what we do to determine reaction enthalpies via heats of formation and Hess's law. Because remember, enthalpy and Gibbs energy are both a state functions.
I realize this is very crude, but here's a figure demonstrating what I mean. It is two Morse potentials for A-B and B-C, where B-C has a lower energy than A-B. You can see that if I wanted the reaction to occur, I could completely break A-B (an input energy of "10", and then form B-C, release an energy of "13", for a net energy gain of "3". However, if I start at the minimum of the blue line and go right by stretching the A-B bond out, and simultaneously form the B-C bond by following the red line also to the right, there is a "crossing" point, where A and C are both partially bonded to B. Following the reaction in this way is following the reaction coordinate. A more realistic path is shown in green, which would be akin to both of these processes happening truly spontaneously. At the maximum of the green line is the transition state. As I keep following it right, you can see that I continue to "lose" energy from continuing to pull apart the A-B bond. However this is more than compensated by the formation of B-C, which simultaneously releases potential energy. Again, very crude figure and totally made up numbers, but maybe it helps convey the idea.
Please also bear in mind that once the transition state is reached, it is not mandatory that the reaction proceed to generation of products. The transition state also represents the mid-point in the reverse reaction as well. All reaction dynamics are governed by statistics. The state of the system once equilibrium has been reached is determined by the relative energies of the reactants and products, which is why the Gibbs energy change is directly related to the equilibrium constant. The time it takes to reach equilibrium is related to (among other things) the energy of the transition state, which governs the probability that two colliding molecules at a certain temperature will have enough kinetic energy to form the transition state. The shape of the potential energy surface also can be used to predict what vibrational modes lead to formation of the activated complex and what kinds of collision trajectories are most likely to bring about a successful reaction. However these topics are getting rather far afield.
NB: I notice a minor error in the figure. The "red" line legends should be "B + C to B-C", not "B-C to B + C".