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How to prove this inequality $$\left(a+\frac{1}{2} \left(a b-\sqrt{a^2-1} \sqrt{b^2-1}\right)\right)^{3/4}-\frac{\sqrt{3} \cos\left[\frac{3 (\pi -t)}{4}\right]}{2 \left(\frac{1}{2}+b\right)^{1/4}}-(a+\cos t)^{3/4}\ge 0$$ for $a,b\ge 1$ and $t\in[0,\pi]$. It can be noticed that for $a=b$ and $t=\pi/3$ the quantity is equal to zero.

This inequality would imply some sharp form of M. Riesz conjugate function theorem.

How to prove this inequality $$\left(a+\frac{1}{2} \left(a b-\sqrt{a^2-1} \sqrt{b^2-1}\right)\right)^{3/4}-\frac{\sqrt{3} \cos\left[\frac{3 (\pi -t)}{4}\right]}{2 \left(\frac{1}{2}+b\right)^{1/4}}-(a+\cos t)^{3/4}\ge 0$$ for $a,b\ge 1$ and $t\in[0,\pi]$. It can be noticed that for $a=b$ and $t=\pi/3$ the quantity is equal to zero.

How to prove this inequality $$\left(a+\frac{1}{2} \left(a b-\sqrt{a^2-1} \sqrt{b^2-1}\right)\right)^{3/4}-\frac{\sqrt{3} \cos\left[\frac{3 (\pi -t)}{4}\right]}{2 \left(\frac{1}{2}+b\right)^{1/4}}-(a+\cos t)^{3/4}\ge 0$$ for $a,b\ge 1$ and $t\in[0,\pi]$. It can be noticed that for $a=b$ and $t=\pi/3$ the quantity is equal to zero.

This inequality would imply some sharp form of M. Riesz conjugate function theorem.

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MathArt
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One specific inequality

How to prove this inequality $$\left(a+\frac{1}{2} \left(a b-\sqrt{a^2-1} \sqrt{b^2-1}\right)\right)^{3/4}-\frac{\sqrt{3} \cos\left[\frac{3 (\pi -t)}{4}\right]}{2 \left(\frac{1}{2}+b\right)^{1/4}}-(a+\cos t)^{3/4}\ge 0$$ for $a,b\ge 1$ and $t\in[0,\pi]$. It can be noticed that for $a=b$ and $t=\pi/3$ the quantity is equal to zero.