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[tex]2\sqrt{n+1}-\sqrt{n}-\sqrt{n+2} >= 0 [/tex]
- Thread starter ahorn
- Start date
First off, I'm assuming that you are working with only n >= 0, because for negative values of n, you get a complex number. You need to remember to always explicitly declare for which values of n the proof is valid. That said, I'd start by rearranging the terms, like so:
If \(\displaystyle 2\sqrt{n+1}-\sqrt{n}-\sqrt{n+2}\ge 0\) for \(\displaystyle n\in \left[0,\infty \right]\) then \(\displaystyle 2\sqrt{n+1}\ge \sqrt{n}+\sqrt{n+2}\) for \(\displaystyle n\in \left[0,\infty \right]\)
Now that the left side of the equation has only one term under a square root, what if you squared both sides? Try working from there. The proof will come together with a bit more manipulation.
If \(\displaystyle 2\sqrt{n+1}-\sqrt{n}-\sqrt{n+2}\ge 0\) for \(\displaystyle n\in \left[0,\infty \right]\) then \(\displaystyle 2\sqrt{n+1}\ge \sqrt{n}+\sqrt{n+2}\) for \(\displaystyle n\in \left[0,\infty \right]\)
Now that the left side of the equation has only one term under a square root, what if you squared both sides? Try working from there. The proof will come together with a bit more manipulation.
Method of proof?
Hi ksdhart. Can you please tell me the method of proof? I usually separate the left-hand side (LHS) and the right-hand side (RHS) and work on those two sides separately. Is there a way of doing this, or is the proof simply based on observation?
If we suppose that the inequality is true (merely as an exercise), then
\(\displaystyle 4(n+1) \ge n + 2\sqrt{n(n+2)} + n + 2\\
\Leftrightarrow 2n + 2 \ge 2\sqrt{n^2+2n}\\
\Leftrightarrow 4n^2+8n+4 \ge 4(n^2+2n) \qquad \text{(squaring again)}\\
\Leftrightarrow 4 \ge 0\)
Which doesn't yield anything.
First off, I'm assuming that you are working with only n >= 0, because for negative values of n, you get a complex number. You need to remember to always explicitly declare for which values of n the proof is valid. That said, I'd start by rearranging the terms, like so:
If \(\displaystyle 2\sqrt{n+1}-\sqrt{n}-\sqrt{n+2}\ge 0\) for \(\displaystyle n\in \left[0,\infty \right]\) then \(\displaystyle 2\sqrt{n+1}\ge \sqrt{n}+\sqrt{n+2}\) for \(\displaystyle n\in \left[0,\infty \right]\)
Now that the left side of the equation has only one term under a square root, what if you squared both sides? Try working from there. The proof will come together with a bit more manipulation.
Hi ksdhart. Can you please tell me the method of proof? I usually separate the left-hand side (LHS) and the right-hand side (RHS) and work on those two sides separately. Is there a way of doing this, or is the proof simply based on observation?
If we suppose that the inequality is true (merely as an exercise), then
\(\displaystyle 4(n+1) \ge n + 2\sqrt{n(n+2)} + n + 2\\
\Leftrightarrow 2n + 2 \ge 2\sqrt{n^2+2n}\\
\Leftrightarrow 4n^2+8n+4 \ge 4(n^2+2n) \qquad \text{(squaring again)}\\
\Leftrightarrow 4 \ge 0\)
Which doesn't yield anything.
Last edited:
D
Deleted member 4993
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Hi ksdhart. Can you prove that the inequality is false for complex numbers? Also, please tell me the method of proof. I usually separate the left-hand side (LHS) and the right-hand side (RHS) and work on those two sides separately. Is there a way of doing this, or is the proof simply based on observation?
If we suppose that the inequality is true (merely as an exercise), then
\(\displaystyle 4(n+1) \ge n + 2\sqrt{n(n+2)} + n + 2\\
\Leftrightarrow 2n + 2 \ge 2\sqrt{n^2+2n}\\
\Leftrightarrow 4n^2+8n+4 \ge 4(n^2+2n) \qquad \text{(squaring again)}\\
\Leftrightarrow 4 \ge 0\)
Which doesn't yield anything.
It does .... think about it...
Better yet .... assume the other direction:
\(\displaystyle 4(n+1) \lt n + 2\sqrt{n(n+2)} + n + 2\\\)
Now you have reductio ad absurdum!!
Proof
Consider the statement \(\displaystyle \sqrt{a-1}+\sqrt{a+1} < 2\sqrt a \).
In order to prove that this statement is true, it suffices to prove that \(\displaystyle \left(\sqrt{a-1}+\sqrt{a+1}\right)^2<\left(2\sqrt a \right)^2 \) since \(\displaystyle \sqrt{a-1}+\sqrt{a+1} \ge 0 \) and \(\displaystyle 2\sqrt a \ge 0 \)
\(\displaystyle \left(\sqrt{a-1}+\sqrt{a+1}\right)^2=a-1+2\sqrt{a^2-1}+a+1\) and \(\displaystyle \left(2\sqrt a \right)^2 = 4a \).
It is evident that \(\displaystyle 2a+2\sqrt{a^2-1}<2a + 2a\).
Now, let \(\displaystyle a= n+1 \qquad q.e.d.\)
Consider the statement \(\displaystyle \sqrt{a-1}+\sqrt{a+1} < 2\sqrt a \).
In order to prove that this statement is true, it suffices to prove that \(\displaystyle \left(\sqrt{a-1}+\sqrt{a+1}\right)^2<\left(2\sqrt a \right)^2 \) since \(\displaystyle \sqrt{a-1}+\sqrt{a+1} \ge 0 \) and \(\displaystyle 2\sqrt a \ge 0 \)
\(\displaystyle \left(\sqrt{a-1}+\sqrt{a+1}\right)^2=a-1+2\sqrt{a^2-1}+a+1\) and \(\displaystyle \left(2\sqrt a \right)^2 = 4a \).
It is evident that \(\displaystyle 2a+2\sqrt{a^2-1}<2a + 2a\).
Now, let \(\displaystyle a= n+1 \qquad q.e.d.\)