For real numbers 0<x<π2, prove that cos2xcotx+sin2xtanx≥1.
My solution:
First, note that we can rewrite cos2xcotx+sin2xtanx as cos3xsinx+sin3xcosx.
For the domain 0<x<π4, we have cos3x>sin3x,1sinx>1cosx, so by the rearrangement inequality we have:
cos2xcotx+sin2xtanx=cos3xsinx+sin3xcosx≥cos3xcosx+sin3xsinx=cos2x+sin2x=1
By the same token, for the domain π4≤x<π2, we have sin3x>cos3x,1cosx>1sinx, so by the rearrangement inequality we also have:
cos2xcotx+sin2xtanx=sin3xcosx+cos3xsinx≥sin3xsinx+cos3xcosx=sin2x+cos2x=1
Combining the two yields the result.
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