i'd like to prove the following inequality: $$ (\int_{0}^{\infty}e^{-\alpha ((A+t)^b-A^b)} dt )^2 \geq \int_{0}^{\infty}t e^{-\alpha ((A+t)^b-A^b)} dt $$
where $\alpha \geq 0$ (scale parameter), $b \geq 1$ (shape parameter) and A positive reel number (age).
In fact, what i im trying to prove is that, a random variable following a weibull distribution with age A, has a coefficient of variation smaller than 1.
I tried to get a closed forme of these 2 integrals, where the first one is (first moment):
$$ m_1=\int_{0}^{\infty}e^{-\alpha ((A+t)^b-A^b)} dt = \frac{\alpha^{-1/b}}{b}e^{\alpha A^b} \Gamma(\frac{1}{b},\alpha A^b)=\frac{A}{b}U(1,1+\frac{1}{b},\alpha A^b) $$ where $\Gamma$ the upper incomplete gamma function and U the confluent Hypergeometric function;
and the second one(half of the 2nd moment): $$\frac{m_2}{2}=\int_{0}^{\infty}t e^{-\alpha ((A+t)^b-A^b)} dt =\frac{\alpha^{-2/b}}{b}e^{\alpha A^b} \Gamma(\frac{2}{b},\alpha A^b)-A \cdot m_1 $$
However, i still don't see how can we prove this inequality.
Cn anyone give some help or references?
Since the two exponents are identical, we can proceed in a very rough naive but interesting way:
Let's instead prove the following:
$$\left(\int_0^{+\infty} e^{-at}\ dt\right)^2 \geq \int_0^{+\infty} t e^{-at}\ dt$$
Now the integrals are pretty trivial, and we conclude:
$$\frac{1}{a^2} \geq \frac{1}{a^2}$$
If instead of $e^{-at}$ we chose $e^{-at^2}$ thence we'd have
$$\left(\frac{\sqrt{\pi}}{2\sqrt{a}}\right)^2 \geq \frac{1}{2a}$$
$$\frac{\pi}{4a} \geq \frac{1}{2a}$$
Which is of course true.
By a simil method we can extend it to your integral, hence the inequality holds true.