An inequality of riemann zeta function

Question

I need show this inequality

Let $\sigma >1$, show that

$$\zeta(\sigma)\geq \frac{\sigma^2}{(\sigma-1)(2\sigma-1)}$$

Any help is appreciated

Thanks!

Answer 1

The bound seems too weak. Anyways, $\zeta(\sigma)$ is the left Riemann sum for $\displaystyle \int_{1}^{\infty}\frac{\mathrm{d}x}{x^{\sigma}}$ we get $$\zeta(\sigma)\ge \displaystyle \int_{1}^{\infty}\frac{\mathrm{d}x}{x^{\sigma}}+\frac{1}{2} =\frac{1}{\sigma-1}+\frac{1}{2}\ge \frac{\sigma^2}{(\sigma-1)(2\sigma-1)}$$ where the penulitmate bound comes from estimating the excess by areas of corresponding triangles, since the function is clearly convex. Hope I was clear enough.