Inequality Integral who involves logarithms and exponentials

Question

Show that: $$\int_1^e (x^2+1)\log^n(x) dx \leq \frac{2}{e^{n+1}} \int_1^e e^x\cdot x^n dx$$

My approach: I tried using Cauchy-Swartz but not work. I tried $\log(x) \leq x, \forall x \geq 1$ but still not work. Any ideas? :/ I'm stuck.

Answer 1

Note that we have the inequalities $$\log(x)\le x/e$$ and $$2e^{x-1}\ge (x^2+1)$$ And we are done!

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To show that $\log(x)\le x/e$, we use the inequality $\log(x)\le x-1$. Letting $y=x/e$, we find that

$$\begin{align} \log(x)&=\log(ey)\\\\ &=\log(y)+1\\\\ &\le (y-1)+1\\\\ &=y\\\\ &=x/e \end{align}$$

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To show that $2e^{x-1}\ge (x^2+1)$, we write $x^2+1=2+2(x-1)+(x-1)^2$. Then, observe that

$$\begin{align} 2e^{x-1}&=\sum_{k=0}^\infty \frac{2(x-1)^k}{k!}\\\\ &=2+2(x-1)+(x-1)^2+\underbrace{\sum_{k=3}^\infty \frac{2(x-1)^k}{k!}}_{\ge 0}\\\\ &\ge x^2+1 \end{align}$$