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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 $$\frac{(4n+3)\sqrt{n}+6\sqrt{n+1}}{6}\leq \frac{(4n+7)\sqrt{n+1}}{6}$$

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  • $\begingroup$ You'll need to take less of a leap on that inequality in line 1. Can you think of something else to do? $\endgroup$ – Ian Coley Sep 3 '13 at 5:17
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    $\begingroup$ Are you really asked to use induction? $\endgroup$ – André Nicolas Sep 3 '13 at 5:19
  • $\begingroup$ Yes, it asks for a prove by induction before I can use the inequality to estimate the sum. $\endgroup$ – please delete me Sep 3 '13 at 5:23
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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.

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  • $\begingroup$ Thank you very much! I noticed that when ever doing induction involving inequalities, it is best to remove any unnecessary terms. Simplifying the expression and leaving only what is necessary to prove makes it much easier. $\endgroup$ – please delete me Sep 3 '13 at 5:37
  • $\begingroup$ @user2357: This is why I asked do you really need to do induction. It is a straightforward non-induction manipulation. $\endgroup$ – André Nicolas Sep 3 '13 at 5:39
  • $\begingroup$ @user2357: You’re very welcome. $\endgroup$ – Brian M. Scott Sep 3 '13 at 5:39
  • $\begingroup$ @André: But this is the induction step of the OP’s argument. $\endgroup$ – Brian M. Scott Sep 3 '13 at 5:40
  • $\begingroup$ @AndréNicolas: $\left ( \frac{4n+3}{6} \right )\sqrt{n}+\sqrt{n+1}\leq \left ( \frac{4n+7}{6} \right )\sqrt{n+1}$ was the third step of induction because the original question involves a series. Is there another way without using induction? $\endgroup$ – please delete me Sep 3 '13 at 5:43

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