<s>[INST] Prove the inequality $1\le\int_1^4 \frac{1}{1+\sqrt(x)} \,dx$ I want to prove the inequality:
$$1\le\int_1^4 \frac{1}{1+\sqrt{x}} \,dx$$
This is my attempt:
The domain is $(0,\infty)$ and the range is $[0,1]$
So, $\frac{1}{1+\sqrt{x}} \ge \frac{1}{1+\sqrt{9}}=\frac{1}{4}$ (used a value for x=9)
So, $1\ge\int_1^4 \frac{1}{1+\sqrt{x}} \,dx \ge \int_1^4 \frac{1}{4} \,dx = \frac{3}{4}$, but this is not a proof. I don't think this is the way I should calculate this inequality
Can anyone help me where I got it wrong? Am I on the right track? Or am I way wrong?
Many thanks:)
 [/INST] The idea is correct but you picked $x=9$ instead of $x=4$. We have for $x\in [1,4]$
$$\frac{1}{1+\sqrt{x}}\geq \frac{1}{1+\sqrt{4}}=\frac{1}{1+2}=\frac{1}{3}$$
Then
$$\int_1^4 \frac{1}{1+\sqrt{x}}dx\geq \int_1^4 \frac{1}{3}dx =\frac{3}{3}=1$$
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