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It appears that

$$\sin(\log(x+1)+1)>\frac{41}{39}-\frac{x}{7}$$

for $x\ge 1$.

However, the bound is extremely tight, with the minimum of $LHS-RHS$ being about $6.6\times10^{-5}$. How can I prove the inequality analytically?

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We have that $\sin(\log(x+1)+1)$ is a convex function for $x\geq 3$, hence it is sufficient to prove the inequality on the interval $[1,3]$ then prove that, if $$ f(x)=\sin(\log(x+1)+1),\qquad g(x)=\frac{41}{39}-\frac{x}{7}$$ and $x_0$ is the point for which $f'(x_0)=-\frac{1}{7}$, we have $f(x_0)>g(x_0)$.

Numerically, neither of these two parts are difficult to prove. We have $x_0 = 5.8311\ldots $.

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