Where is your evidence of three times greater pressure difference under the wing than above it?

Okay. One more time for GrahamH.

Here is the video's location that I mentioned:

https://www.youtube.com/watch?v=3_WgkVQWtno If you fast forward to 4 min: 00secs and pause it you should be able to observe the smooth laminar flow of the smoke trails under the wing. On the far right of the picture the atmospheric pressure is 1bar (1 x atmospheric pressure) and the smoke trails are evenly spaced, Now look at the smoke trails immediately under the trailing edge of the wing. You will observe that three smoke trails have been compressed into the space that one smoke trail occupies on the far right (about the size of the cursor on your screen)

Now that hasn't been achieved by the smoke trails going faster. Indeed we know that the air below the wing travels slower than the air above it (thousands of references available for this). The air has therefore been compressed to one third of its original volume.

Boyle's Law tells us that if you compress a quantity of gas (e.g. air) to one third of its volume then the pressure of that gas will be increased threefold. The pressure under the trailing edge of the wing must, therefore be 3bar (3 x atmospheric pressure).

Now let's look at the smoke trails above the wing. Here we observe that the smoke trails have dissipated in an area of low pressure due to the steep angle of attack. If that low pressure was actually a total vacuum then the pressure would be zero bar (0 x atmospheric pressure), which is one bar less than atmospheric pressure. It doesn't matter, by the way, whether the vacuum is 1mm thick, 1 metre thick or 100 kilometres thick, the reduction in pressure acting downwards on the wing is still 1 bar.

So the maximum possible (hypothetical) lift generated by the upper wing surface is equal to the force generated by one atmosphere, while, as we have seen, the observed upward force generated by air compression under the wing is equal to three times atmospheric pressure.

I hope I've made that clear. Thanks for your input.