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I have trouble figuring out how to show that

$\frac{\sqrt{n+1}}{(n+1)^{2}+1}<\frac{\sqrt{n}}{n^{2}+1},\quad\forall n\geq1$.

I have solved several inequalities before, but I can't prove this one.


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Hint The sequence is decreasing if the derivative is negative. – Daryl Nov 14 '12 at 5:26
Unfortunately, I cannot use derivatives. Only inequalities are allowed. – Hubble Nov 14 '12 at 5:28
up vote 2 down vote accepted

Your inequality is equivalent, after multiplying through by the product of denominators, and squaring the (positive) sides, to $$n((n+1)^2+1)^2>(n+1)(n^2+1)^2.$$ In this if the sides are expanded one arrives at the inequality $$3n^4+6n^3+6n^2+3n-1>0,$$ clearly true for positive $n.$

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This is equivalent to $\sqrt{1+1/n} < ((n+1)^2+1)/(n^2+1)$.

Since $\sqrt{1+1/n} < 1+1/(2n)$ and $((n+1)^2+1)/(n^2+1) = 1+(2n+1)/(n^2+1)$, this is true if $1/(2n) < (2n+1)/(n^2+1)$ or $4n^2+2n > n^2+1$, which is true for $n \ge 1$.

Note that more than this is true: Let's try to find the largest value of $c > 1$ such that $\sqrt{1+c/n} < ((n+1)^2+1)/(n^2+1)$.

As before, $\sqrt{1+c/n} < 1+c/(2n)$ so we want $c/(2n) < (2n+1)/(n^2+1)$ or $n^2+1 < (2/c)n(2n+1) =(4/c)n^2+2/c$.

For $c = 4$ this is false, but it is true for large enough $n$ for any $c < 4$. To see this, let $4/c = 1+d$ where $d > 0$. Then we want $n^2+1 < (1+d)n^2 + (1+d)/2$ or $d n^2 > 1-(1+d)/2 = (1-d)/2$ or, assuming $d < 1$ (this is true for all $n$ if $d \ge 1$), $n^2 > (1-d)/(2d) = 1/(2d) -1/2$. Thus $n > \sqrt{1/2d}$ works.

So, there is no largest value of $c$ such that $\sqrt{1+c/n} < ((n+1)^2+1)/(n^2+1)$ (at least with this elementary method), but the inequality is true for any $c < 4$ for all large enough $n$.

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Let $$ f(x)=\frac{\sqrt{x}}{x^2+1}. $$ Then it is easy to check $$ f'(x)=\frac{-3x^2+1}{2\sqrt{x}(x^2+1)}\le 0 $$ for $x>1$. It means that $f(x)$ is strictly decreasing. so $f(n+1)< f(n)$ or $$ \frac{\sqrt{n+1}}{(n+1)^2+1} < \frac{\sqrt{n}}{n^2+1}. $$

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