<s>[INST] Proving Inequality with Square Roots and Algebraic Manipulation Question:
Edit: Original question is: Estimate $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{10000}$ to nearest hundred.
I used $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\geq \int_{0}^{n}\sqrt{x}dx$ and $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n}\leq \frac{4n+3}{6}\sqrt{n}$, which I need to prove.
So prove that: $$\left ( \frac{4n+3}{6} \right )\sqrt{n}+\sqrt{n+1}\leq \left ( \frac{4n+7}{6} \right )\sqrt{n+1}$$
Attempt:
$$\left ( \frac{4n+3}{6} \right )\sqrt{n}+\sqrt{n+1}= \frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+3)\sqrt{n+1}+6\sqrt{n+1}}{6}= \frac{(4n+9)\sqrt{n+1}}{6}$$
But $$\frac{(4n+9)\sqrt{n+1}}{6}\geq \frac{(4n+7)\sqrt{n+1}}{6}$$
How can I prove that $$\frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+7)\sqrt{n+1}}{6}$$
 [/INST] Your desired inequality
$$\frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+7)\sqrt{n+1}}{6}$$
is equivalent to 
$$(4n+3)\sqrt{n}\le(4n+1)\sqrt{n+1}\;.$$
Since we’re dealing with non-negative numbers here, this is equivalent to
$$n(16n^2+24n+9)\le(n+1)(16n^2+8n+1)\;.$$
If you expand that, you’ll see that it’s true.
</s>