Why doesn't a siphon send water upwards? The input side of a THIN siphon tube dips in a tank of water. The output side of the tube is several times FATTER AND SHORTER and ends far above the input end. Because the output side is several times fatter, it has more weight of water than in the input side, so it will pull the water downwards and out of the output end, even though the output end is above the input end. If the output end goes to a second tank which is higher than the first tank then water will continuously flow upwards from the lower tank to the higher tank without requiring any external energy source.
We always learn that a siphon works because the weight of the water in the longer output end is more than that in the shorter input end. Here I am achieving the same thing with the short and FAT output end, which has more weight of water. Obviously, the water will flow faster in the thin end and slower in the fat end to make this work.
I am not looking for the answer that this is impossible due to the conservation of energy or that perpetual motion is impossible. I want to know why this is impossible in terms of forces and pressures etc.
 A: I had not figured it out before, but have figured it out now, so am posting an answer instead of deleting it. The standard explanation of the siphon is that the fluid in the longer end of the siphon has more weight than the shorter end, so it moves downward, "pulling" the fluid in the shorter end along with it, and an example is shown with a large and small weight tied by a string over a pulley. But that is only a kid's short cut explanation of a siphon, and works perfectly if the tube is constant diameter.
However fluids do not move due to just the weight, they move due to pressure. And for a constant diameter siphon, if we divide the weight on both sides by the same cross section areas, we get a higher pressure on the longer side and a smaller pressure at the shorter side. So the fluid on the longer side still moves down and the fluid on the shorter side still moves up.
In the question which was asked, the shorter side had a higher cross-section area and also more weight of fluid, but when we divide by the higher cross-section area, the pressure was smaller, because after dividing the weight by the cross-section areas, the cross-section area is canceled out, we are left with the formula for pressure = $\rho g h$, which depends just on the height, and is independent of the cross section area. So the pressure in the shorter fatter side will be smaller just because the height of the fluid is smaller, and the pressure in the long thin side is higher because the height is larger.
Therefore, the fluid will always move out of the longer end and into the shorter end. Making the shorter end fatter will not change this because the cross-section area was cancelled out when we determined the pressure.
A: The flow in the wider diameter part is slower.
Though it may seem logical to compare the weight difference of the wide and narrow column, the difference in diameter must also be considered.
Equal volumes must flow in equal time, therefor a tube 10x in diameter will have 1/10th the flow rate.  The longer tube has more weight per area (psi), so its FxD is greater than the short tubes FxD for equal volume flow.
This can be reasoned by putting the ends of the tubes at equal fluid level.  There will be no flow.
The ancient Egyptians had a fairly ingenious solution to moving water uphill, a solid weight on on end of a pivoting fulcrum and a bucket on the other.  Archemides developed the rotating threaded screw.  Powered impeller pumps generally do the job today.
Plants move water "uphill" by osmosis, but only very slowly and in limited quantities.
Here is a possible test apparatus (for proof):

A: Changing the cross-section of the tube doesn't help.  The extra area of the larger tube is supported by the atmosphere pushing up on the bottom of the column of water.  The larger you make the tube, the more area you have for the atmosphere to push up on it.

