$\forall a>0$, prove that $\int_{0}^{a} x^{-1/2} \operatorname e^{2x}\operatorname d\!x\le \frac{3}{2}a^{1/6}\operatorname e^{2a}$

Question

Let a>0

Prove with Hölder's Inequality that;

$$\int\limits_{x=0}^{x=a} x^{-1/2} \operatorname e^{2x}\operatorname d\!x\le \frac{3}{2}a^{1/6}\operatorname e^{2a}$$

this was a question, asked at the exam today. and I couldn't solve it.

I mean if you find some p and q and consider the second integral $(\int_{0}^{a} (e^{2x})^q)^{1/q}$

you always get $(\frac{e^{2aq}-1}{2q})^{1/q}$. and you cannot simplify it.

Answer 1

Hint: Let $q=3$ and use the obvious inequality

$$(e^{6a}-1)^{1/3}\lt e^{2a}$$