How to prove the following integral inequality problem

I have a question about an inequality I am trying to integrate. $$\int_0^1 (1 - x^a)^b ~dx \ge \left(1-\frac{1}{a+1}\right)\frac{1}{b^{\frac{1}{a}}},$$ where $$a, b \ge 1$$ are integers.

Any assistance with this problem would be appreciated.

My attempt: $$\int_0^1 (1-x^{a})^b dx= \frac{1}{a}\int_0^1 (1-t)^bt^{\frac{1}{a}-1}dt$$$$=\frac{1}{a}B\left(\frac{1}{a},b+1\right)dt$$$$=\frac{1}{a}\frac{\Gamma\left(\frac{1}{a}\right)\Gamma(b+1)}{\Gamma\left(\frac{1}{a}+b+1\right)}$$$$\ge \left(1-\frac{1}{a+1}\right) \frac{1-\epsilon}{d^{\frac{1}{a}}},\text{ works only when b is large enough}.$$

Here I use the facts that $$\frac{1}{a}\Gamma\left(\frac{1}{a}\right)\ge 1-\frac{1}{a+1}$$ and $$\lim_{b\rightarrow \infty}\frac{\Gamma(b+1)}{\Gamma(b+1+\frac{1}{a})b^{\frac{1}{a}}}=1$$

Could someone prove the inequality for grneral integers $$a,b$$?

• I would think to use here Bernoulli inequality but I don't see how do you get the factor $1/b^{1/a}$? Commented Jun 25 at 17:50
• AH, maybe first integrate by parts, and after one integration by parts to use Bernoulli inequality. Commented Jun 25 at 17:52
• @MathematicalPhysicist I guess the following might be related to $b^{1/a}$. A result of Gamma function says that $\lim_{b\rightarrow \infty} \frac{\Gamma(b+\frac{1}{a})}{\Gamma(b)n^{\frac{1}{a}}}=1$ (if we take $x^a=t$ in the integral, then Gamma function appears). Commented Jun 25 at 17:58

The desired inequality is written as $$\int_0^1 (1 - x^a)^b b^{1/a}\,\mathrm{d} x + \frac{1}{1 + a} \ge 1.$$
We have \begin{align*} &\int_0^1 (1 - x^a)^b b^{1/a}\,\mathrm{d} x + \frac{1}{1 + a}\\ ={}& \frac{1}{a} \int_0^1 (1 - t)^b t^{1/a - 1} b^{1/a}\,\mathrm{d} x + \frac{1}{1 + a} \tag{1}\\ \ge {}& \frac{1}{a} \int_0^{1/b} (1 - t)^b t^{1/a - 1} b^{1/a}\,\mathrm{d} x + \frac{1}{1 + a}\\ ={}& \frac{1}{a} \int_0^1 (1 - u/b)^b u^{1/a - 1}\,\mathrm{d} u + \frac{1}{1 + a}\tag{2}\\ \ge{}& \frac{1}{a} \int_0^1 (1 - u) u^{1/a - 1}\,\mathrm{d} u + \frac{1}{1 + a}\tag{3}\\ ={}& 1. \end{align*} Explanations:
(1): Use the substitution $$t = x^a$$;
(2): Use the substitution $$t = u/b$$;
(3): Use the Bernoulli inequality $$(1+u)^r \ge 1 + ur$$ for all $$u > -1$$ and $$r \ge 1$$.
A generalization of the Gautschi's inequality states that (see https://math.stackexchange.com/a/2089967/326159) $$\frac{1}{sn^{s-1}} \leq \frac{\Gamma(n+1)}{s\Gamma(n+s)} = \frac{\Gamma(n+1)}{\Gamma(n+s+1)}, \qquad 0 \leq s\leq 1, \qquad n \in \mathbb{Z}^+.$$
With your notation, let $$b=n$$ and $$s = \tfrac{1}{a}$$, so $$\int_0^1(1-x^a)^b\text{d}x = \frac{\Gamma(\tfrac{1}{a})\Gamma(b+1)}{a\Gamma(b+\tfrac{1}{a}+1)} \geq \frac{\Gamma(\tfrac{1}{a})}{b^{1/a}} \geq \left(1-\frac{1}{1+a}\right)\frac{1}{b^{1/a}}.$$