# Upper bounds for $\frac{x_1}{1+x_1^2} + \frac{x_2}{1 + x_1^2 + x_2^2} + \cdots + \frac{x_n}{1 + x_1^2 + x_2^2 + \cdots + x_n^2}$

Problem: Let $$x_1, x_2, \cdots, x_n$$ ($$n\ge 2$$) be reals. Find upper bounds for $$\frac{x_1}{1+x_1^2} + \frac{x_2}{1 + x_1^2 + x_2^2} + \cdots + \frac{x_n}{1 + x_1^2 + x_2^2 + \cdots + x_n^2}.$$ There is also the following Ji Chen's estimation (mentioned in the link below): $$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\dotsb+\frac{x_n}{1+x_1^2+x_2^2+\dotsb+x_n^2}<\sqrt{n}-\dfrac{\ln{n}}{2\sqrt{n}}.\tag{1}$$

Question: How to prove the bound (1)? Can we obtain better upper bounds?

Any comments and solutions are welcome and appreciated.

Edit(2022/02/22): The problem can be rephrased as follows:
Let $$c_1 = 1/2$$ and $$c_{k + 1} = g(c_k), k \ge 1$$ where $$g(c) = \frac18\sqrt{-2c^4 + 40c^2 + 16 + 2c(c^2 + 8)\sqrt{c^2 + 8}}.$$ Find the upper bounds of $$c_n$$.

Some bounds:

1. IMO ShortList 2001, algebra problem 3, see: https://artofproblemsolving.com/community/c6h17449p119163 $$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\dotsb+\frac{x_n}{1+x_1^2+x_2^2+\dotsb+x_n^2}<\sqrt{n}. \tag{2}$$ zhaobin@AoPS gave a very nice proof for (2): \begin{align*} &\mathrm{LHS}^2\\ \le\ & n\left(\frac{x_1^2}{(1+x_1^2)^2}+\frac{x_2^2}{(1+x_1^2+x_2^2)^2}+\dotsb+\frac{x_n^2}{(1+x_1^2+x_2^2+\dotsb+x_n^2)^2}\right)\\ \le\ & n\Big(\frac{x_1^2}{1\cdot (1+x_1^2)}+\frac{x_2^2}{(1+x_1^2)(1+x_1^2+x_2^2)}+ \frac{x_3^2}{(1+x_1^2+x_2^2)(1+x_1^2+x_2^2+x_3^2)}\dotsb\Big)\\ \le\ & n\Big(1 - \frac{1}{1+x_1^2} + \frac{1}{1+x_1^2} - \frac{1}{1+x_1^2+x_2^2} + \frac{1}{1+x_1^2+x_2^2} - \frac{1}{1+x_1^2+x_2^2+x_3^2}\cdots\Big)\\ \le\ & n\left(1 - \frac{1}{1+x_1^2+x_2^2 + \cdots + x_n^2}\right)\\ <\ & n. \end{align*}

My attempt:

I found the following relation: Let $$y_k = \frac{x_{k+1}}{\sqrt{1+x_1^2}}, k = 1, 2, \cdots, n-1$$ and we have $$\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2} = \frac{x_1}{1 + x_1^2} + \left(\sum_{m=1}^{n-1} \frac{y_m}{1 + \sum_{k=1}^m y_k^2}\right)\frac{1}{\sqrt{1+x_1^2}}.$$ By this, if we have $$\sum_{m=1}^{n-1} \frac{x_m}{1 + \sum_{k=1}^m x_k^2} \le F(n-1)$$ on $$\mathbb{R}^{n-1}$$ for some function $$F(\cdot)$$, then we have $$\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2} \le g(F(n-1))$$ on $$\mathbb{R}^n$$ where $$g(c) \triangleq \max_{x\in \mathbb{R}} \frac{x}{1+x^2} + c \frac{1}{\sqrt{1+x^2}}.$$ Remark: $$g(c)$$ admits a closed form: $$g(c) = f\left(\sqrt{\tfrac{2}{c^2 + 2 + c\sqrt{c^2 + 8}}},\ c\right) = \frac{\sqrt{\frac{2}{c\, \sqrt{c^2 + 8} + c^2 + 2}}}{1 + \frac{2}{c\, \sqrt{c^2 + 8} + c^2 + 2}} + \frac{c}{\sqrt{1 + \frac{2}{c\, \sqrt{c^2 + 8} + c^2 + 2}}}$$ where $$f(x, y) = \frac{x}{1+x^2} + \frac{y}{\sqrt{1+x^2}}$$.

We immediately have the following results:

1. The maximum of $$\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2}$$ is given by $$\underbrace{g\circ g \circ \cdots \circ g}_{n-1} \left(\frac{1}{2}\right).$$

Indeed, denote the maximum of $$\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2}$$ by $$M(n)$$, and we have $$M(n) = g(M(n-1))$$. Also, $$M(1)$$ is equal to the maximum of $$\frac{x_1}{1+x_1^2}$$ which is $$1/2$$. The desired result follows.

1. A upper bound for $$\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2}$$ is $$\sqrt{n}$$.

We use mathematical induction. When $$n=1$$, it is true. Assume $$\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2} < \sqrt{n}$$. We need to prove that $$g(\sqrt{n}) < \sqrt{n+1}$$. It is true.

Similarly, we can obtain $$\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2} \le \sqrt{n} - \frac{1}{2\sqrt{n}}$$.

However, this does not work for $$\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2} < \sqrt{n}-\frac{\ln{n}}{2\sqrt{n}}$$ since $$g(\sqrt{n}-\frac{\ln{n}}{2\sqrt{n}}) < \sqrt{n+1}-\frac{\ln{n+1}}{2\sqrt{n+1}}$$ is not true.

• For $a,x\in(0,1)$ define : $$\frac{x^{\frac{1}{2}}}{x+1+a}=f(x)$$ then the maximum occurs at $x=1+a$ . RiverLi can you confirm ? Feb 17, 2022 at 17:30
• Is this helpful for my question? Feb 18, 2022 at 1:14
• I think it works for $a,x\in(0,\infty)$ .As you can guess we get a series in $n$.$$0.5+\sum_{k=1}^{n}\frac{1}{2\sqrt{k+\frac{1}{2}}}$$ .See wolframalpha.com/… Feb 18, 2022 at 10:38
• Can you use your result to obtain any bounds better than $\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2} \le \sqrt{n} - \frac{2}{\sqrt{n}}$? Feb 18, 2022 at 10:49
• As you pointed out there is a recursive part .I guess that (we need to show it) the maximum is obtained with the formula in my second comment . If true I answer your question with a yes ! It implies the zeta function . Feb 18, 2022 at 11:11

If i am correctly understanding the question, the inequality is not in the target of the question, but rather starting with it, we obtain a recursive way to get the best margin on its constant size, when $$n$$ variables are allowed, and the question wants us to analyze the asymptotics of these best margins.

Let us get some better formula for $$g$$ first: \begin{aligned} g(c) &\triangleq \max_{t\ge 0} \frac t{1+t^2} + c \frac 1{\sqrt{1+t^2}}\qquad(t=\tan u)\\ &= \max_{0\le u< \pi/2} \frac {\tan u}{1+\tan^2 u} + c \frac 1{\sqrt{1+\tan^2 u}}\\ &= \max_{0\le u< \pi/2} \sin u\cos u + c \cos u\qquad(x=\sin u\ ,\ y=\cos u)\\ &= \max_{\substack{0\le x,y\le 1\\x^2+y^2=1}} xy+cy \\ &= \frac 1{4\sqrt 2} \cdot \frac{5c^2 + 3c\sqrt{c^2 + 8} + 4} {\sqrt{c^2 + c\sqrt{c^2 + 8}+2}}\ . \end{aligned} This is because the associated Lagrange multiplier system for the function $$F(x,y;\lambda)= xy+cy-\lambda(x^2+y^2-1)$$, \left\{ \begin{aligned} 0 &= y - 2\lambda x\\ 0 &= x+c - 2\lambda y\\ 1 &= x^2 + y^2 \end{aligned} \right. has the solution $$\displaystyle x_*=\frac 14(\sqrt{c^2+4}-c)$$, $$\displaystyle y_*=\frac{\sqrt 2}4\cdot\frac{3c + \sqrt{c^2 + 8}}{\sqrt{c^2 + c\sqrt{c^2 + 8} + 2}}$$ in $$[0,1]^2$$.

Since our asymptotic is needed for big values of $$n$$, it is useful to conjugate $$g$$ with the involution $$j$$ defined by $$j(c)=1/c$$. The obtained function $$h=j\circ g\circ j$$ has the formula in a point $$s:=1/c$$ given by $$h(s) = 4\sqrt 2 \cdot s \cdot \frac {\sqrt{1 + \sqrt{1+8s^2}+2s^2}} {5 + 3\sqrt{1+8s^2} + 4s^2} \ .$$ In a picture:$$\require{AMScd}$$ $$\begin{CD} c_1=\frac 12 @>g>> c_2 @>g>> c_3 @>g>> c_4 @>g>> \dots \\ @VjVV @VjVV @VjVV @VjVV \\ s_1=2 @>>h> s_2 @>>h> s_3 @>>h> s_4 @>>h> \dots \end{CD}$$

We want an upper bound for the $$g$$-iterations in $$c_1 = 1/2$$, $$c_{n+1}=g(c_n)$$ of the shape $$c_n\le \sqrt n-(?)$$ with an "ambitious" part $$(?)$$. Using $$h$$ instead, we need a lower bound for the $$h$$-iterations in $$s_1=2$$, $$s_{n+1}=h(s_n)$$ of the shape $$\displaystyle\frac 1{\sqrt n}+(??)\le s_n$$.

It may be useful to see some numerical values:

for k in [1, 10**2, 10**4, 10**6]:
for m in [1, 4, 9]:
n = k*m
val = rec(h, 1, 2, n).n(100)
print('s({:>7}) is approx {}'.format(n, val))


Results:

s(      1) is approx 2.0000000000000000000000000000
s(      4) is approx 0.68006721588682742534047497623
s(      9) is approx 0.39713327155406107196210216858
s(    100) is approx 0.10269804298633590654576884222
s(    400) is approx 0.050417448038696684151142231933
s(    900) is approx 0.033471476463956208613835517759
s(  10000) is approx 0.010004917283275061470414817518
s(  40000) is approx 0.0050007010350304313800943056724
s(  90000) is approx 0.0033335560456167706527492633495
s(1000000) is approx 0.0010000072167319931007495666427
s(4000000) is approx 0.00050000098872864891066981192997
s(9000000) is approx 0.00033333364130678928794916060580


(The above sage code uses functions defined in the postponed code section.) It is natural to conjecture the asymptotic behaviour for the above as $$\tag{*_{\log}} \frac 1{\sqrt n}\left( 1 + \frac{\log n}{2n} \right) \overset !\le s_n$$ immediately after looking at the values of $$s_n$$ for $$n=1,10^2,10^4, 10^6$$. But this is not so simple to show (and type). Here, to illustrate how one can work when only a polynomial shape (and no $$\log$$) is in the game, i will show $$\color{blue}{ \tag{*} \frac 1{\sqrt n}\left( 1 + \frac1n \right) \overset !\le s_n\ , \qquad n\ge 1\ . }$$ The Taylor expansion of $$h$$ around zero is: $$s - \frac{1}{2} s^{3} + \frac{7}{8} s^{5} - \frac{41}{16} s^{7} + \frac{1259}{128} s^{9} - \frac{11351}{256} s^{11} + \frac{227267}{1024} s^{13} +O(s^{15}) % - \frac{2447417}{2048} s^{15} + \frac{222455219}{32768} s^{17} - \frac{2633079659}{65536} s^{19}+\dots \ .$$ In particular, $$h$$ is strictly increasing in a suitable neighborhood of zero. Since we need a lower bound for the $$h$$-iterates, it is enough to work with an increasing version $$f\le h$$ of $$h$$, and obtain for it lower bounds. Of course, the better the Taylor approximations of $$f$$ and $$h$$ match, the better lower bounds will be obtained. Here, for illustration (and easy typing) $$f$$ is: $$f(s) = s\left(1-\frac 12s^2+\frac 34 s^4\right)\ ,$$ Graphical comparison on some bigger scale:

Of course, the dark red $$f$$ becomes rapidly bigger when the $$s^7$$-part in $$h$$ gets sensible compared to the $$s^5$$-part in the Taylor expansion of the difference $$h-f$$, but the approximation works good on the interval $$[0,1/5]$$, where the picture for $$(h-f)(s)/s^5$$ is:

Our strategy to proceed is now as follows. Use computer software to check the claimed inequality $$(*)$$ for $$n$$ between one and $$k=100$$, when $$s_k\approx 0.1026980429863359065457688422\dots$$ lands inside the interval $$[0, 1/5]$$, and in this interval $$f\le h$$. Then consider the sequence $$(t_n)$$ with $$t_1=s_1$$, $$t_2=s_2$$, ... $$t_{100}=s_{100}$$ and $$t_{n+1}=f(t_n)\text{ for }n\ge 100\ .$$ Then inductively $$t_n\le s_n$$, and so it is enough to show $$\color{blue}{ \tag{\dagger} \frac 1{\sqrt n}\left( 1 + \frac1n \right) \overset !\le t_n\ , \qquad n\ge 1\ . }$$ We would then like to show: $$\color{navy}{ \tag{\maltese} \frac 1{\sqrt {n+1}}\left( 1 + \frac1{n+1} \right) \le f\left(\frac 1{\sqrt n}\left( 1 + \frac1n \right)\right) }$$ so that we can conclude inductively (for $$n\ge 100$$): $$\color{navy}{ \frac 1{\sqrt {n+1}}\left( 1 + \frac1{n+1} \right) \overset{(\maltese)}\le f\left(\frac 1{\sqrt n}\left( 1 + \frac1n \right)\right) } \le f(t_n)=t_{n+1} \le s_{n+1} \ .$$ Then $$\color{navy}{\maltese}$$ is successively equivalent to: \begin{aligned} \frac 1{\sqrt {n+1}}\left( 1 + \frac1{n+1} \right) &\overset{(\maltese)}\le f\left(\frac 1{\sqrt n}\left( 1 + \frac1n \right)\right)\ , \\ \frac 1{\sqrt {n+1}}\left( 1 + \frac1{n+1} \right) &\le \frac 1{\sqrt n}\left( 1 + \frac1n \right) \left[\ 1 - \frac 1{2n}\left( 1 + \frac1n \right)^2 + \frac 3{4n^2}\left( 1 + \frac1n \right)^4 \ \right] \ , \\ 1 &\le \left(1+\frac 1n\right)^{1/2} \cdot \frac{1 + \frac1n}{1 + \frac{1/n}{(n+1)/n}} \left[\ 1 - \frac 1{2n}\left( 1 + \frac1n \right)^2 + \frac 3{4n^2}\left( 1 + \frac1n \right)^4 \ \right] \ , \\ 1 &\le \left(1+x\right)^{1/2} \cdot \frac{1 + x}{1 + \frac x{1+x}} \left[\ 1 - \frac 12 x(1 + x)^2 + \frac 34 x^2( 1 + x)^4 \ \right] \ , \\ &\qquad\qquad\text{ where }x =\frac 1n\ , \\ 1 &\le 1 + \frac{3}{8} x^{2} + \frac{1}{2} x^{3} + \frac{1163}{128} x^{4} + \frac{23}{32} x^{5} + \frac{16415}{1024} x^{6} \\ &\qquad\qquad - \frac{93}{4} x^{7} + \frac{1595667}{32768} x^{8} - \frac{398267}{4096} x^{9} + \frac{50975389}{262144} x^{10} +\dots \ , \end{aligned} and the plot the above function of $$x$$ shows that we have indeed a true inequality for $$x=\frac 1n\in[0,1]$$. (We need in fact only values of $$x$$ less $$1/100$$.)

This concludes $$(*)$$.

$$\square$$

Note: The above estimation should be enough for most purposes. In fact, the function $$g$$ used in the recursion is useful only when related with the given inequality. This inequality may be of interest for small, yes, very small values of $$n$$ like $$1,2,3,4,5$$. But i cannot imagine that we really want to do something with the starting inequality for numbers like $$n=100$$ (and up to hundred we can use numerical approximations), or even investigate asymptotics related to $$g$$-recursions. But ok, the question was related to the asymptotics, asking for ideas to attack the situation, and above we have a decent way to get with low effort a good shape for it. to get more, the effort and the obtained result should be in some ballance, i do not see the reason to invest more effort.

Note: To estimate from below in the same manner also having the logarithmic part, i.e. something like: $$\frac 1{\sqrt {n+1}}\left( 1 + \frac{\log (n+1)}{2(n+1)} \right) \le h\left(\frac 1{\sqrt n}\left( 1 + \frac{\log n}{2n} \right)\right)\ ,\qquad\text{ i.e.}$$ does not work inductively.

Sage code used:

def h(s):
y = sqrt(1+8*s^2)
return 4*sqrt(2) * s * sqrt(1 + y + 2*s^2)/(5 + 3*y + 4*s^2)

def f(s):
return s*(1 - 1/2*s^2 + 3/4*s^4)

def myid(s):
return s

var('s');
taylor(h(s), s, 0, 13)

ph1 = plot([h, myid, f], [0,2]
, ymin=0, ymax=2
, color=['red', 'navy', 'darkred'])
ph1.show()

ph2 = plot(lambda s: (h(s) - f(s))/s^5, [0, 0.3], ymin=0, ymax=0.3)
ph2.show()

def rec(fun, k, xk, n):
"""Compute the recursion for the function fun
starting from the value xk in k, return the value in n.
"""
a = xk.n(digits=200)
for j in range(k, n):
a = fun(a).n(digits=200)
return a


We can compare in the sage interpreter $$(s_n)$$ and $$(t_n)$$ for some values of the index $$n$$ the transparent corresponding terms to see the similarity in the "asymptotic character" of the two sequences:

sage: s100 = rec(h, 1, 2, 100)
sage: s10000 = rec(h, 1, 2, 10000)
sage: s1000000 = rec(h, 1, 2, 1000000)
sage: t100 = s100
sage: t10000 = rec(f, 100, s100, 10000)
sage: t1000000 = rec(f, 100, s100, 1000000)

sage: s100.n(100)
0.10269804298633590654576884222
sage: t100.n(100)
0.10269804298633590654576884222

sage: s10000.n(100)
0.010004917283275061470414817518
sage: t10000.n(100)
0.010004356266589258071961130982

sage: s1000000.n(100)
0.0010000072167319931007495666427
sage: t1000000.n(100)
0.0010000060808712179410156533239


Check for $$(*)$$ and $$n$$ between $$1$$ and $$30$$, and $$s_{30}$$ is already in the interval $$[0,\ 0.2]$$ where $$f\le h$$:

ok = True    # optimistic hypothesis about (*)
a = 2.
for n in [1..30]:
bound = ( 1/sqrt(n) * ( 1 + 1/n ) ).n(50)
if bound > a:
ok = false
print(f'*** INVALID INEQUALITY FOR n = {n}')
break
print(f'n = {n} :: OK :: bound = {bound} is less sn ~ {a.n(50)}')
a = h(a).n(200)


Results:

n = 1 :: OK :: bound = 2.0000000000000 is less sn ~ 2.0000000000000
n = 2 :: OK :: bound = 1.0606601717798 is less sn ~ 1.1362522106647
n = 3 :: OK :: bound = 0.76980035891950 is less sn ~ 0.83618545460737
n = 4 :: OK :: bound = 0.62500000000000 is less sn ~ 0.68006721588683
n = 5 :: OK :: bound = 0.53665631459995 is less sn ~ 0.58282683401858
n = 6 :: OK :: bound = 0.47628967220784 is less sn ~ 0.51569070797355
n = 7 :: OK :: bound = 0.43195939772483 is less sn ~ 0.46613903659504
n = 8 :: OK :: bound = 0.39774756441743 is less sn ~ 0.42781519344335
n = 9 :: OK :: bound = 0.37037037037037 is less sn ~ 0.39713327155406
n = 10 :: OK :: bound = 0.34785054261852 is less sn ~ 0.37190890468059
n = 11 :: OK :: bound = 0.32892146681211 is less sn ~ 0.35073116948516
n = 12 :: OK :: bound = 0.31273139581105 is less sn ~ 0.33264559483497
n = 13 :: OK :: bound = 0.29868472104435 is less sn ~ 0.31698186443143
n = 14 :: OK :: bound = 0.28635133062045 is less sn ~ 0.30325443114657
n = 15 :: OK :: bound = 0.27541214906364 is less sn ~ 0.29110232382475
n = 16 :: OK :: bound = 0.26562500000000 is less sn ~ 0.28025117409299
n = 17 :: OK :: bound = 0.25680242650906 is less sn ~ 0.27048841709749
n = 18 :: OK :: bound = 0.24879683041749 is less sn ~ 0.26164660646525
n = 19 :: OK :: bound = 0.24149024617954 is less sn ~ 0.25359189427448
n = 20 :: OK :: bound = 0.23478713763748 is less sn ~ 0.24621589447692
n = 21 :: OK :: bound = 0.22860921834247 is less sn ~ 0.23942981942445
n = 22 :: OK :: bound = 0.22289165800814 is less sn ~ 0.23316017814100
n = 23 :: OK :: bound = 0.21758025814651 is less sn ~ 0.22734556929131
n = 24 :: OK :: bound = 0.21262931794993 is less sn ~ 0.22193425540024
n = 25 :: OK :: bound = 0.20800000000000 is less sn ~ 0.21688230376460
n = 26 :: OK :: bound = 0.20365906341273 is less sn ~ 0.21215214453894
n = 27 :: OK :: bound = 0.19957787083098 is less sn ~ 0.20771144009125
n = 28 :: OK :: bound = 0.19573160209406 is less sn ~ 0.20353218949284
n = 29 :: OK :: bound = 0.19209862570040 is less sn ~ 0.19959001265570
n = 30 :: OK :: bound = 0.18865999202956 is less sn ~ 0.19586357316850

• Thank you very much for the detailed answer! (+1) I need some time to check it. Feb 15, 2022 at 23:53
• Thanks for providing another way to investigate the $g$-iterations. Your $g(c)$ is equal to mine, also is equal to $\frac18\sqrt{-2c^4 + 40c^2 + 16 + 2c(c^2 + 8)\sqrt{c^2 + 8}}$. Feb 16, 2022 at 2:41
• By the way, as the aim of the question, I hope to see a proof of Ji Chen's estimation (1). My approach can obtain $\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2} \le \sqrt{n} - \frac{2}{\sqrt{n}}$ which is far from Ji Chen's one. Feb 16, 2022 at 2:48

Extended comment :

Using what I say to user RiverLi I found a better upper bound it seems we have for $$n\geq 11$$ :

$$\frac{x_1^{0.5}}{1+x_1} + \frac{x_2^{0.5}}{1 + x_1 + x_2} + \cdots + \frac{x_n^{0.5}}{1 + x_1 + x_2+ \cdots + x_n}<0.5\left(\frac{x_1^{0.5}}{x_1+1}+\frac{1}{2\sqrt{1+x_{1}+x_{2}}}+\cdots+\frac{1}{2\sqrt{1 + x_1 + x_2 + \cdots + x_n}}+\sqrt{n}-\frac{\ln\left(n\right)}{2\sqrt{n}}\right)$$

Hope it helps !

Edit :

I want to make clear some details for user RiverLi :

First of all we introduce the function for $$a,x \in (0,\infty)$$ :

$$f(x)=\frac{x^{0.5}}{x+1+a}$$

We have a lemma :

Lemma :

The maximum of $$f(x)$$ is obtained for the abscissa $$x=1+a$$

Proof :

$$f'(x)=\frac{a+1-x}{2\sqrt{x}\left(x+1+a\right)^{2}}$$

So now we have the recursive part :

The first term is $$x_1$$ so we build the left hand side as follow :

For $$n=2$$ we have :

$$\frac{x_1^{0.5}}{x_1+1}+\frac{x_2^{0.5}}{x_2+x_1+1}$$

The part $$\frac{x_2^{0.5}}{x_2+x_1+1}$$ have the form of $$f(x)$$ with $$x_1=a$$

So the maximum is reached for $$x_2=1+x_1$$ so it's a one variable inequality

Now the case $$n=3$$

$$\frac{x_1^{0.5}}{x_1+1}+\frac{x_2^{0.5}}{x_2+x_1+1}+\frac{x_3^{0.5}}{x_2+x_1+x_3+1}$$

As in the case $$n=2$$ we have $$x_2=1+x_1$$ and $$x_3=1+x_2+x_1=2x_1+2$$ as said it's a one variable inequality :

In the general case we have :

$$\frac{x_1^{0.5}}{1+x_1} + \frac{x_2^{0.5}}{1 + x_1 + x_2} + \cdots + \frac{x_n^{0.5}}{1 + x_1 + x_2 + \cdots + x_n}\leq \frac{x_1^{0.5}}{x_1+1}+\frac{1}{2\sqrt{1+x_{1}+x_{2}}}+\cdots+\frac{1}{2\sqrt{1 + x_1 + x_2 + \cdots + x_n}}$$

Now it's seems easier but I haven't the knowledge to continue .

Edit 20/02/2022 :

For $$x_i\geq 1$$ with the result above we can choose $$x_i=1$$ remains to show :

$$0.5+\sum_{k=1}^{n}\frac{1}{2\sqrt{k+2}}-\left(\sqrt{n}-\frac{\ln\left(n\right)}{2\sqrt{n}}\right)<0$$

On the other hand we have ($$n\geq 2$$) :

$$0.5+\sum_{k=1}^{n}\frac{1}{2\sqrt{k+2}}-\left(0.5+\int_{0}^{n}\left(\frac{1}{2\sqrt{x+2}}\right)dx\right)<0$$

Now it's really easy .

• It seems not true for $n = 50$. Please check it. The iteration is given by $c_1 = 1/2, c_{k+1} = g(c_k), k = 1, 2, \cdots$ where $g(c) = \frac18\sqrt{-2c^4 + 40c^2 + 16 + 2c(c^2 + 8)\sqrt{c^2 + 8}}$ (or use dan_fulea's formula). Feb 18, 2022 at 13:03
• @RiverLi see my edits Thanks again for your patience . Feb 18, 2022 at 19:21
• It seems not true since $\frac{x_1}{x_1^2+1}+\left(\frac{\sqrt{2}}{4}+\frac{\sqrt{4}}{8}+\frac{\sqrt{8}}{16}+\cdots\right)\frac{1}{\sqrt{1+x_1^2}}$ is bounded (notice $\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2}$ should be in order $\sqrt{n}$). You can check it numerically. Feb 18, 2022 at 23:30
• @RiverLi I go too far now it seems true . Feb 19, 2022 at 8:40
• I want a form of $\sum_{m=1}^n \frac{x_m}{1 + \sum_{k=1}^m x_k^2} \le f(n)$. Is your result helpful? Feb 19, 2022 at 9:12

Structure and systematics are well known to Mathematics.

So start of with a simple infinite sequence. Start with 1 constant.

The problem is then

$$\sum_{k=1}^{n}\frac{x_{k}}{1+\sum_{l=1}^{l}x_{l}^{l}}$$

I hope there is an agreement to this.

This is equal to

$$-1+EulerGamma+PolyGamma(0,2+n)$$

in full generality. I wrote for better identification purpose the trivial names in full. $$n$$ is still free. $$n$$ is a natural integer as assumed.

A short plot and all the given arguments are gone:

The upper bound of the problem is matched for all n bigger than 0.

As mentioned at the start this can be done for all reals

$$\sum_{k=1}^{n}\frac{a}{1+\sum_{l=1}^{l}a_{k}^{2}}=$$

$$\frac{-PolyGamma(0,1+\frac{1}{a^2})+-PolyGamma(0,1+\frac{1}{a^2}+n)}{a}$$

That are infinitely many examples following the first one.

The complications arise to small values of $$a$$ and in the definition problems of the PolyGamma function.

Constant sequences do not seem too impressive but give a nice introduction to the so general problem.

The next steps are, to do this for the most important sequences.

$$\sum_{k=1}^{n}\frac{\frac{1}{k}}{1+\sum_{l=1}^{l}(\frac{1}{k})^{2}}=$$

$$-1+EulerGamma+PolyGamma(0,2+n)$$

So the same as our starting point.

Look at an example of a divergent sequence for example the integers itself:

$$\sum_{k=1}^{n}\frac{k}{1+\sum_{l=1}^{l}k^{2}}=$$

The result can not be given in a closed form. It has complicated rootsums. I took a screen in a famous CAS for an idea:

So this is seemingly the basis for all possible finite sequences that can be invented and done into the sum of quotients of the sequence at the summation index divided by one plus the finite sum over the sequence up to the summation index.

Zoom into the critical region compared to the function of the problem is

So limits and borders are that of PolyGamma functions. This works fine because of the definition of the PolyGamma functions. I personally have several problems open. First I can not find another source for this definition of the PolyGamma function. There might be more correction to the $$\sqrt(n)$$ upper border/limit. I did work out that. The artifact vanishes for more plot points. There are many more sequences that can be tried and stay conform. This can be extended into the complexes and follow still the PolyGamma function. PolyGamma is defined in well known formula collection like Abramovitch, Stegun, and CAS.

• Thanks. What do you mean "The problem is then $\sum_{k=1}^{n}\frac{x_{k}}{1+\sum_{l=1}^{l}x_{l}^{l}}$"? (Particularly, $\sum_{l=1}^{l}x_{l}^{l}$) Feb 10, 2022 at 15:52