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pikachu975

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This is probably gonna be a dumb question but what's the symbol on the LHS, e.g. (something - a^2)?
 

fan96

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First rearrange the inequality:









Because , we have





Importantly, because , this explicit formula for shows that and thus .

Therefore the inequality is equivalent to



Finally,

 

CM_Tutor

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Importantly, because , this explicit formula for shows that and thus .

Therefore the inequality is equivalent to



Finally,

I am not certain that the last part of this is valid.

I agree that we have and that , but as a consequence the term



must be non-zero:







Similar reasoning allows us to conclude that



satisfies



And, while it may be true that , this evidence does not support any conclusion stronger than

Consider some values:

Case 1: Both near zero









Case 2: Both near one









Case 3: Both near middle









Case 4: One near zero, one near one









Looking at these four cases, we see that is typically not close to zero, and that the sum does stay below . The conclusion based on and is not reflected in any of these cases, however.
 

fan96

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My bad, I didn't check that last step correctly.

This may not quite be within the MX2 syllabus, but if you differentiate the expression



twice, as a function of and then as a function of , it can be shown that there is exactly one point in the unit square that is a stationary point in both cases.

(The derivative actually only needs to be taken once, as and are symmetric in the expression.)

This point can be shown to be a local maximum by analysing the two derivatives.

The function attains the value of at this point, so must be its maximum value over the unit square. (It's actually a global maximum, but that isn't necessary to show.)
 

CM_Tutor

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One slight cheat would be to say that the equation that we have is symmetrical in and and thus that the max / min likely lies on the diagonal . We would then need to examine the function:



in the domain . We would then find:



which is zero when and so the only stationary point in the domain corresponds to , as @fan96 stated.



Proving that corresponds to a maximum.

Substituting will show that and thus completes the proof of the inequality.

I don't plan to type it out, but I have checked, and when , we get



and



and thus get

 
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Paradoxica

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Random observation I want to make. Let



Then the problem reduces to maximising



subject to the constraint



and A,B,C are acute angles.

I don't know if this has a nice geometric interpretation, but if anyone has one, feel free to share it.

Edit: oh this was the solution described two posts above
 
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