An inequality for a concave function $f(x)=x^{p/2}$

Question

Assume that $p\in(1,2]$, $a,b\ge 1$, $b\le -\frac{1}{2} \left(\cos\frac{\pi }{p}+\sec\frac{\pi }{p}\right)$, and $t\in[0,\pi]$. How to prove this inequality $$\left(\frac{a+\cos t}{b+\cos\frac{\pi }{p}}\right)^{p/2}\geq \left(\frac{a+\left(a b-\sqrt{a^2-1} \sqrt{b^2-1}\right) \cos\frac{\pi }{p}}{b+\cos\frac{\pi }{p}}\right)^{p/2}+\frac{\cos\frac{p t}{2} \sin\frac{\pi }{p}}{b+\cos\frac{\pi }{p}},$$

Answer 1

This is an answer to the question as it was originally formulated by the OP.

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For $p=1$, $a=1$, $b=2$ the left-hand-side equals $\sqrt{1+\cos t}$, but the right-hand-side equals $i$, so the inequality does not hold.