Tighter logarithmic inequality

Question

There is a well-known lower bound for $$ x\log{1+x\over x}\geq {x\over1+x} $$ for $x\geq0$. I know a tighter lower bound on the same domain $$ x\log{1+x\over x}\geq{2x\over1+2x}\geq {x\over1+x}. $$ It can be proved by Jensen's inequality. Do you know something tighter?

Answer 1

If any tighter lower bound will suffice, here you go with a ratio of quadratics: $$x \log \frac{x+1}x \ge \frac{3 x \,(2x+1)}{6x^2+6x+1} \ge \frac{2x}{2x+1}$$

Proving the rightmost inequality is easy with cross multiplication to get $\iff x \ge 0$. For the left inequality, you need to show for $x > 0$, $$ f(x)= \log \left(1+\tfrac1x \right) - \frac{6x+3}{6x^2+6x+1} \ge 0$$ which is evident from $\displaystyle f'(x) = -\frac1{x(x+1)(6x^2+6x+1)^2} < 0$ and $\lim_{x \to \infty} f(x) = 0$.

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P.S. You can keep getting better approximations using larger polynomials $\dfrac{p(x)}{q(x)}$ - one way is to ensure that this matches in limits ($x \to 0, x \to \infty$) with your LHS, then get as many terms of the Taylor series (or as $x \to \infty$) to match as you want. You may also want to google or read up on Pade approximants.