krypticlemonjuice
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- 2023
In part (ii), you showed thatView attachment 39983
Can someone please do this question and show me their working?
View attachment 39984
I have no idea what this working out means, especially the first line
why is a, b, c being less than 1 a problem tho cos ii is true for all positive a, b, cIn part (ii), you showed that
so long as a, b, and c are real and positive.
where x, y, and z are suitably chosen positive reals. Putting these into the equation from part (ii):
and thus you get the solution's inequation (1), except in x, y, and z.
Similar substitutions yield the other three inequations, and they sum to give the result required.
The problem with this approach is that the substitutions collectively require a, b, and c to be less than 1.