Given x≥y≥z≥π/12, x+y+z=π/2, find the maximum and minimum of cosxsinycosz.
I tried using turn siny to cos(x+z), and Jensen Inequality, but filed. Please help. Thank you.
*p.s. I'm seeking for a solution without calculus.
Answer
Let P=cosxsinycosz=cosz2[2cosxsiny]=cosz2[sin(x+y)−sin(x−y)]≤cosz2⋅sin(x+y)
So P≤cosz⋅cosz2=14(1+cos2z)≤14[1+cos2⋅π12]=2+√38
Above equality hold when sin(x−y)=0⇒x=y and given x+y=π2−z and x≥y≥z≥π12
And for max(P), We must have z=π12 and x=y=5π24
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