I have found that $\left|\frac{\zeta'(x)}{\zeta^2(x)}\right|\leq \frac{1}{x-\frac{1}{2}}$ for all real $x$ such that $x>1$ seems to be true. I have plotted the inequality and got this inequality holding. Can someone help me to prove it?
Edit: It seems to be too hard to prove it so if you know any similar inequalities can you share their proofs?
In the range $1<x<1.5$, the known bounds $$-\frac{\zeta'(x)}{\zeta(x)}<\frac{1}{x-1}<\zeta(x)$$ imply that $$-\frac{\zeta'(x)}{\zeta^2(x)}<1<\frac{1}{x-1/2}.$$ In the range $1.5\leq x<1.7$, we have by monotonicity that $$-\frac{\zeta'(x)}{\zeta(x)}\leq-\frac{\zeta'(1.5)}{\zeta(1.5)}<1.6<\frac{\zeta(1.7)}{1.7-0.5}<\frac{\zeta(x)}{x-1/2}.$$ In the range $1.7\leq x<2$, we have by monotonicity that $$-\frac{\zeta'(x)}{\zeta(x)}\leq-\frac{\zeta'(1.7)}{\zeta(1.7)}<1<\frac{\zeta(2)}{2-0.5}<\frac{\zeta(x)}{x-1/2}.$$ In the range $2\leq x<2.6$, we have by monotonicity that $$-\frac{\zeta'(x)}{\zeta(x)}\leq-\frac{\zeta'(2)}{\zeta(2)}<0.6<\frac{\zeta(2.6)}{2.6-0.5}<\frac{\zeta(x)}{x-1/2}.$$ In the range $2.6\leq x<3$, we have by monotonicity that $$-\frac{\zeta'(x)}{\zeta(x)}\leq-\frac{\zeta'(2.6)}{\zeta(2.6)}<0.3<\frac{\zeta(3)}{3-0.5}<\frac{\zeta(x)}{x-1/2}.$$ In the range $3\leq x$, the known bound $\psi(r)<1.04 r$ for the second Chebyshev function implies that $$-\frac{\zeta'(x)}{\zeta^2(x)}<-\frac{\zeta'(x)}{\zeta(x)}=x\int_2^\infty\frac{\psi(r)}{r^{x+1}}\,dr<\frac{2.08x}{x-1}2^{-x}<\frac{1}{x-1/2}.$$ The quoted bounds can be found in Delange (1987), Corollary 1.14 of Montgomery-Vaughan: Multiplicative number theory I, and Theorem 12 of Rosser-Schoenfeld (1962).
