I'm studying the gravitationnal field produced by the Sun at a point at the surface of the Earth and at the center of the planet. I've reduced the situation to a geometry problem but I'm stucked with it. I've made a Geogebra simulation and I've seen that it works but I can't proove it correctly. I'm pretty sure it's easy to solve but I can't despite many tries...
Let ABC be a triangle. AC is the longest side. How can we simply prove that $AC^{\,2}\geq(AB-BC)^{\,2}$ ?
Any tips would be very very appreciated because I'm quite desperate with that ^_^'
You use the triangle inequality. If $AB,BC$ and $AC$ are the three sides of a triangle, then $AB+BC\ge AC$. The inequality is then manipulated as follows:
$$AB+BC-BC \ge AC-BC \Longrightarrow AB \ge AC-BC$$
Of course, the longest side of your triangle is $AC$, which allows us to write:
$$AC \ge AB \ge AC-BC$$
Now suppose that $BC\le AB$. Clearly, $AB \ge AB-BC \ge 0.$ Inserting this into the above inequality yields:
$$AC \ge AB \ge AC-BC \ge AB-BC$$
Now, we actually want the terminal terms of the inequality; the middle ones don't matter, so we can ignore them. This gets us $$AC\ge AB-BC \Longrightarrow AC^2\ge (AB-BC)^2$$ as desired. QED.
Note that it could be the case that $BC\ge AB$ instead of the other way around. No worries, the proof works exactly the same, except you'd begin by subtracting $AB$ in the starting inequality instead of $BC$. Either way, you get the same result, all as a consequence of the triangle inequality.