Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

\[ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2} < \sqrt{n}. \] \[ \iff \left(\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\right) ^2< n. \] but we have: \[ \left(\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\right)^2 \le n\left(\frac{x^2_1}{(1+x_1^2)^2} + \frac{x^2_2}{(1+x_1^2 + x_2^2)^2} + \cdots +\frac{x^2_n}{(1 + x_1^2 + \cdots + x_n^2)^2}\right) \] \[ \le n\left(\frac{x^2_1}{1\cdot (1+x_1^2)} + \frac{x^2_2}{ (1+x_1^2)(1+x_1^2 + x_2^2)} + \cdots +\frac{x^2_n}{(1 + x_1^2 + \cdots + x_{n-1}^2)(1 + x_1^2 + \cdots + x_n^2)}\right)=n\left(1-\frac{1}{1 + x_1^2 + \cdots + x_n^2}\right)<n \]

We prove the following equivalent statement: for $ k>0$, \[ \sum_{j=1}^n \frac{x_j}{k^2 + \sum_{i=1}^j x_i^2} \le \sqrt{n}/k, \] with equality only when $ n=0$. For proof, we induct on $ n$. Our base case, $ n=0$, is a trivial equality. Now, suppose the inequality holds for $ n$. Then by inductive hypothesis, \[ \sum_{j=1}^{n+1} \frac{x_j}{k^2 + \sum_{i=1}^j x_i^2} = \frac{x_1}{k^2 + x_1^2} + \sum_{j=2}^{n+1} \frac{x_j}{k^2 + x_1^2 + \sum_{i=2}^{n+1}x_j^2} \le \frac{x_1}{k^2 + x_1^2} + \frac{\sqrt{n}}{\sqrt{k^2+x_1^2}} . \] Setting, $ t = \text{Arcsin} \left( x_1/ \sqrt{k^2 + x_1^2} \right)$, we have \[ \frac{x_1}{k^2 + x_1^2} + \frac{\sqrt{n}}{\sqrt{k^2+x_1^2}} = (\cos t \sin t + \sqrt{n} \cos t)/k \le (\lvert \sin t \rvert + \sqrt{n} \cos t)/k, \] with equality only if $ \cos t = \pm 1$. A classic application of Cauchy-Schwarz then yields \[ (\lvert \sin t \rvert + \sqrt{n} \cos t)/k \le (1+n)^{1/2}(\sin^2 t + \cos^2 t)^{1/2}/k = \sqrt{n+1}/k . \] Since our equality cases never coincide, our desired strict inequality thus holds by induction. $ \blacksquare$

By Cauchy we have \[\sum_{cyc}a_ib_i\le\sqrt{\sum_{cyc}a_i^2}\sqrt{\sum_{cyc}b_i^2}\] with \[a_i=1, b_i=\dfrac{x_i}{1+\displaystyle\sum_{k=1}^{i}x_{k}^2}\] and it follows that \[\text{LHS}\le\text{RHS}\sqrt{\sum_{cyc}b_i^2}\] The rest immediately follows from $b_1^2\le\frac{x_1^2}{1+x_1^2}=1-\frac{1}{1+x_1^2}. \Box$

Please explain these two steps, \[ \left(\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\right)^2 \le n\left(\frac{x^2_1}{(1+x_1^2)^2} + \frac{x^2_2}{(1+x_1^2 + x_2^2)^2} + \cdots +\frac{x^2_n}{(1 + x_1^2 + \cdots + x_n^2)^2}\right) \] \[ \le n\left(\frac{x^2_1}{1\cdot (1+x_1^2)} + \frac{x^2_2}{ (1+x_1^2)(1+x_1^2 + x_2^2)} + \cdots +\frac{x^2_n}{(1 + x_1^2 + \cdots + x_{n-1}^2)(1 + x_1^2 + \cdots + x_n^2)}\right) \]

velraman wrote: \[ \left(\frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots +\frac{x_n}{1 + x_1^2 + \cdots + x_n^2}\right)^2 \le n\left(\frac{x^2_1}{(1+x_1^2)^2} + \frac{x^2_2}{(1+x_1^2 + x_2^2)^2} + \cdots +\frac{x^2_n}{(1 + x_1^2 + \cdots + x_n^2)^2}\right) \] Cauchy Schwarz on the pairs $\left(1, \frac{x_1^2}{(1+x_1^2)^2}\right), \left(1, \frac{x_2^2}{(1+x_1^2+x_2^2)^2}\right), ..., \left(1, \frac{x_n^2}{(1+x_1^2+x_2^2)^2}\right)$ velraman wrote: \[ n\left(\frac{x^2_1}{(1+x_1^2)^2} + \frac{x^2_2}{(1+x_1^2 + x_2^2)^2} + \cdots +\frac{x^2_n}{(1 + x_1^2 + \cdots + x_n^2)^2}\right) \le n\left(\frac{x^2_1}{1\cdot (1+x_1^2)} + \frac{x^2_2}{ (1+x_1^2)(1+x_1^2 + x_2^2)} + \cdots +\frac{x^2_n}{(1 + x_1^2 + \cdots + x_{n-1}^2)(1 + x_1^2 + \cdots + x_n^2)}\right) \] Because $1+x_1^2 \ge 1, 1+x_1^2+x_2^2 \ge 1+x_1^2, ..., 1+x_1^2+x_2^2+...+x_{n-1}^2+x_n^2 \ge 1+x_1^2+x_2^2+...+x_{n-1}^2$, so $(1+x_1^2)^2 \ge 1(1+x_1^2), (1+x_1^2+x_2^2)^2 \ge (1+x_1^2)(1+x_1^2+x_2^2), ..., (1+x_1^2+x_2^2+...+x_{n-1}^2+x_n^2)^2 \ge (1+x_1^2+x_2^2+...+x_{n-1}^2)(1+x_1^2+x_2^2+...+x_{n-1}^2+x_n^2)$.

Is this solution right? First,we take $ x_{n}=\frac{tan \theta_{n}}{\Pi_{j=1}^{n-1} cos \theta_{j}}=\frac{sin \theta_{n}}{\Pi_{j=1}^{n} cos \theta_{j}} , \theta_{i} \in [0,\frac{\pi}{2}) $ for every $ i<n $ so, $ LHS=K(n,\theta_1,\theta_2,\dots,\theta_{n})=\Sigma_{i=1}^{n} sin \theta_{i}\Pi_{j=1}^{i} cos \theta_{j} $ Consider that $ g(\theta)=f(r,\theta)=(r+sin \theta)cos \theta $ => $ K(n,\theta_1,\theta_2,\dots,\theta_{n})=f(f(\dots f(f(0,\theta_{n}),\theta_{n-1})\dots),\theta_1) $ then,we will find the maximum of $ g(\theta) $ for $ r \geq 0 $ let $ g'(\theta)=cos2\theta-rsin\theta=0 $ => $ sin \theta=\frac{-r+\sqrt{r^2+8}}{4} , cos \theta=\frac{\sqrt{8-2r^2+2r\sqrt{r^2+8}}}{4} $ $ g''(\theta)=-2sin2\theta-rcos\theta<0 $ so, $ g(\theta) \leq \frac{3r+\sqrt{r^2+8}}{4}\times\frac{\sqrt{8-2r^2+2r\sqrt{r^2+8}}}{4} $ now,we will prove $ \frac{3r+\sqrt{r^2+8}}{4}\times\frac{\sqrt{8-2r^2+2r\sqrt{r^2+8}}}{4}<\sqrt{r^2+1} $ <=> $ (5r^2+4+3r\sqrt{r^2+8})(4-r^2+r\sqrt{r^2+8})<64(r^2+1) $ <=> $ r(r^2+8)\sqrt{r^2+8}<r^4+12r^2+24 $ <=> $ r^2(r^2+8)^3<(r^4+12r^2+24)^2 $,which is true so, $ g(\theta)<\sqrt{r^2+1} $ => $ LHS<\sqrt{f(\dots f(f(0,\theta_{n}),\theta_{n-1})\dots,\theta_2)^2+1}<\dots<\sqrt{f(0,\theta_{n})^2+n-1}<\sqrt{n} $

Uhhh,i find an another solution from me... let $ x_{1}=tan y_{1} , x_{i}=\frac{tan y_{i}}{\Pi_{k=1}^{i-1} cos y_{i}} $ n=1 is trivial assume that n=k-1 is right while n=k: $ LHS< (\sqrt{k-1}+sin y)cos y $ we will prove $ (\sqrt{k-1}+sin y)cos y < \sqrt{k} $ => $ k-(1-x^2)(x^2+2\sqrt{k-1}x+k-1) \geq 0 $ ($ x=sin y $ here) => $ (x^2+\sqrt{k-1}x-1)^2+x^2 \geq 0 $ so we get the proof with MIT

Excuse, is the result 1 the minimal number to be the coefficient of n^1/2?

One friend (Ji Chen) conjectured that \[\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\dotsb+\frac{x_n}{1+x_1^2+x_2^2+\dotsb+x_n^2}<\sqrt{n}-\dfrac{\ln{n}}{2\sqrt{n}}\]holds for any \(\boldsymbol{x}_n\in\mathbb{R}_+^n\).

Another nice problem may be found in here. In addition, two similar problems are achieved: Jì Chén (pers. comm.) wrote: Let \(x_1, x_2, \dotsc, x_n\) be arbitrary real numbers. Prove that \[\frac{1}{1+x_1}+\frac{2}{1+x_1+x_2}+\dotsb+\frac{n}{1+x_1+x_2+\dotsb+x_n}\leq\frac{n}{2}\sqrt{\frac{1}{x_1}+\frac{1}{x_2}+\dotsb+\frac{1}{x_n}}\textnormal{.}\] Gù Gōng (pers. comm.) wrote:

Solution with trig sub suggested by awang11. Let \[\theta_i = \arctan \frac{x_i}{\sqrt{1+x_1^2+x_2^2+\cdots+x_{i-1}^2}}.\]Then we need to show \[\sqrt{n}>\sum_i \frac{x_i}{1+x_1^2+\cdots+x_i^2} = \sum_i \frac{\tan\theta_i}{\sec^2\theta_i}\cdot\frac{1}{\sqrt{1+x_1^2+\cdots+x_{i-1}^2}}=\sum_i \sin\theta_i\cos\theta_i\cos\theta_{i-1}\cdots\cos \theta_1.\]Now, we induct on $n$. The base case is trivial, as \[\sin\theta_1\cos\theta_1=\frac 12 \sin(2\theta_1)\le \frac 12.\]Then in the induction step it is sufficient to show \[\sqrt{n-1}\cos\theta+\frac 12\sin(2\theta)\le \sqrt{n}.\]Let $\cos\theta=x$ and write \[\sqrt{n-1}\cos\theta+\frac 12\sin(2\theta)=x\sqrt{n-1}+x\sqrt{1-x^2}.\]By squaring, the problem amounts to \[x^2(n-1)+2\sqrt{n-1}x^2\sqrt{1-x^2}+x^2(1-x^2)\le n\iff\]\[2\sqrt{n-1}x^2\sqrt{1-x^2}\le n-nx^2+x^4.\]Let $x^2=y\le1$ so we need to show \[2\sqrt{n-1}y\sqrt{1-y}\le n-ny+y^2\iff 2(n-1)y^2(1-y)\le n^2-2n^2y+n^2y^2+2ny^2-2ny^3+y^4\iff\]\[0\le n^2-2n^2y+n^2y^2+2y^2-2y^3+y^4 = (y^2-y)^2+y^2+(ny-1)^2.\]This is trivial, so we are done.

$LHS^2<=n(\frac{x_1^2}{(1+x_1^2)^2}+\cdots +\frac{x_n^2}{(1+x_1^2+x_2^2+\cdots +x_n^2)^2})$,then say $y_i=x_i^2$,we only need to prove that:$\frac{y_1}{(1+y_1)^2}+\cdots +\frac{y_n}{(1+y_1+y_2+\cdots +y_n)^2}<1$. Clearly,for all $i$,we have $\frac{y_i}{(1+y_1+y_2+\cdots +y_i)^2}<\frac{y_i}{(1+y_1+y_2+\cdots +y_{i-1})(1+y_1+y_2+\cdots +y_{i-1}+y_i)}=\frac{1}{1+y_1+y_2+\cdots +y_{i-1}}-\frac{1}{1+y_1+y_2+\cdots +y_{i-1}+y_i}$ sum $i$ and we are done.

Assume that $x_1,x_2,\dots,x_n$ are nonnegative. It's enough to show that \[\left(\frac{x_1}{1+x_1^2}\right)^2+\left(\frac{x_2}{1+x_1^2+x_2^2}\right)^2+\dots+\left(\frac{x_n}{1+x_1^2+\dots+x_n^2}\right)^2<1\]so by substituting $1+x_1^2+\dots+x_i^2=y_i$ we see that we want to show \[\frac{y_1-1}{y_1^2}+\frac{y_2-y_1}{y_2^2}+\dots+\frac{y_n-y_{n-1}}{y_n^2}<1,\]where $1\le y_1\le y_2\le \dots \le y_n$. We now use induction. Note that the claim clearly holds for $n=1$ so note by the inductive hypothesis that \[\frac{y_2-y_1}{y_2^2}+\frac{y_3-y_2}{y_3^2}+\dots+\frac{y_n-y_{n-1}}{y_n^2}<\frac{1}{y_1}\]so it suffices to show that \[\frac{y_1-1}{y_1^2}+\frac{1}{y_1}\le 1\iff (y_1-1)^2\ge 0\]so we are done. $\blacksquare$

asdf334 wrote: Assume that $x_1,x_2,\dots,x_n$ are nonnegative. It's enough to show that \[\left(\frac{x_1}{1+x_1^2}\right)^2+\left(\frac{x_2}{1+x_1^2+x_2^2}\right)^2+\dots+\left(\frac{x_n}{1+x_1^2+\dots+x_n^2}\right)^2<1\]so by substituting $1+x_1^2+\dots+x_i^2=y_i$ we see that we want to show \[\frac{y_1-1}{y_1^2}+\frac{y_2-y_1}{y_2^2}+\dots+\frac{y_n-y_{n-1}}{y_n^2}<1,\]where $1\le y_1\le y_2\le \dots \le y_n$. We now use induction. Note that the claim clearly holds for $n=1$ so note by the inductive hypothesis that \[\frac{y_2-y_1}{y_2^2}+\frac{y_3-y_2}{y_3^2}+\dots+\frac{y_n-y_{n-1}}{y_n^2}<\frac{1}{y_1}\]so it suffices to show that \[\frac{y_1-1}{y_1^2}+\frac{1}{y_1}\le 1\iff (y_1-1)^2\ge 0\]so we are done. $\blacksquare$ Don’t the y terms need to be squared though

By C-S inequality, \[ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2 + x_2^2} + \cdots + \frac{x_n}{1 + x_1^2 + \cdots + x_n^2} \le n\left(\frac{x_1^2}{(1+x_1^2)^2}+\dots +\frac{x_n^2}{(1 + x_1^2 + \cdots + x_n^2)^2}\right).\]So it suffices to show \[\frac{x_1^2}{(1+x_1^2)^2}+\dots +\frac{x_n^2}{(1 + x_1^2 + \cdots + x_n^2)^2}<1,\]but this is obvious because letting $S_i=1+x_1^2+\dots + x_i^2$ the inequality becomes \[\frac{S_1-1}{S_1^2}+\dots +\frac{S_n-S_{n-1}}{S_n^2}<1.\]Note that this is true because \[\frac{S_1-1}{S_1^2}+\dots +\frac{S_n-S_{n-1}}{S_n^2}< \frac{S_1-1}{S_1}+\frac{S_2-S_1}{S_1S_2}+\dots +\frac{S_n-S_{n-1}}{S_nS_{n-1}}= 1-\frac{1}{S_n}<1\]as desired.

By Cauchy, $$LHS^2\leq n(\frac{x_1^2}{(1+x_1^2)^2}+\frac{x_2^2}{(1+x_1^2+x_2^2)^2}\cdots +\frac{x_n^2}{(1+x_1^2\cdots +x_n^2)^2}),$$so it suffices to show that $$\frac{x_1^2}{(1+x_1^2)^2}+\frac{x_2^2}{(1+x_1^2+x_2^2)^2}\cdots +\frac{x_n^2}{(1+x_1^2\cdots +x_n^2)^2}< 1.$$Let $y_i=x_i^2$, and for the sake of symmetry, let $y_0=1$, and rewrite this as $$\frac{y_0}{y_0^2}+\frac{y_1}{(y_0+y_1)^2}\cdots +\frac{y_n}{(y_0\cdots +y_n)^2}<2.$$Now, we have $$\frac{y_m}{(y_0\cdots +y_m)^2}\leq \frac{1}{(y_0\cdots +y_{m-1})}-\frac{1}{(y_0\cdots +y_m)},$$so we have $$\frac{y_0}{y_0^2}+\frac{y_1}{(y_0+y_1)^2}\cdots +\frac{y_n}{(y_0\cdots +y_n)^2}$$$$\leq 1+\frac{1}{y_0}-\frac{1}{y_0+y_1}+\frac{1}{y_0+y_1}-\frac{1}{y_0+y_1+y_2}+\frac{1}{y_0+y_1+y_2}\cdots -\frac{1}{y_0+y_1\cdots +y_n}<2,$$as desired.

By CBS, \begin{align*} &\phantom{.} \left(\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 + \cdots + x_n^2}\right)^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} + \cdots + \frac{x_n^2}{(1 + x_1^2 + \cdots + x_n^2)^2}\right). \end{align*}Thus, it suffices to prove that \[\frac{x_1^2}{(1 + x_1^2)^2} + \frac{x_2^2}{(1 + x_1^2 + x_2^2)^2} + \cdots + \frac{x_n^2}{(1 + x_1^2 + \cdots + x_n^2)^2} < 1.\]It's easy to see that the LHS is at most \[\le \frac{x_1^2}{1(1 + x_1)} + \frac{x_2^2}{(1 + x_1^2)(1 + x_1^2 + x_2^2)} + \cdots + \frac{x_n^2}{(1 + \cdots + x_{n - 1})(1 +\cdots + x_n)}.\]Notice that for any $S$ and $k$, \[\frac{x_k^2}{S(S + x_k^2)} = \frac{1}{S} - \frac{1}{S + x_k^2}.\]Thus, the above sum telescopes to \[1 - \frac{1}{1 + x_1^2 + \cdots + x_n^2} < 1.\]This completes the proof.

Let $\theta_1 = \cos^{-1} \frac{1}{\sqrt{1 + x_1^2}}$ and for all $i \ge 2$, let \[\theta_i = \cos^{-1} \frac{\sqrt{1 + x_1^2+ \cdots + x_{i - 1}^2}}{\sqrt{1 + x_1^2 + \cdots + x_{i - 1}^2 + x_{i}^2}}.\]We wish to show that \[\sum_{i = 1}^n \frac{x_i}{1 + x_1^2 + \cdots + x_i^2} = \sum_{i = 1}^n \cos\theta_1 \cos\theta_2 \cdots \cos\theta_i \sin \theta_i < \sqrt{n}.\]To prove that $\sum_{i = 1}^n \cos \theta_1 \cdots \cos \theta_i \sin \theta_i < \sqrt{n}$ holds for any $\theta_i$, we will induct on $n$. The base case $n = 1$ is clear because \[\cos \theta_1 \sin \theta_1 = \frac{1}{2} \sin(2\theta_1) < 1.\]Now, onto the inductive step. We can assume that for general $\alpha_i$, \[\cos \alpha_1 \sin \alpha_1 + \cos \alpha_1 \cos \alpha_2 \sin \alpha_2 + \cdots + \cos \alpha_1 \cdots \cos \alpha_{k - 1} \sin \alpha_{k - 1} < \sqrt{k - 1}.\]Multiplying both sides by $\cos \theta_1$, this becomes \begin{align*} &\cos \theta_1 \cos \alpha_1 \sin \alpha_1 + \cos \theta_1 \cos \alpha_1 \cos \alpha_2 \sin \alpha_2 + \cdots + \cos \theta_1\cdots \cos \alpha_{k - 1} \sin \alpha_{k - 1} \\ &< \sqrt{k - 1} \cos \theta_1. \end{align*}Adding $\cos \theta_1 \sin \theta_1 = \frac{1}{2} \sin(2\theta_1)$ to both sides, this is equivalent to \begin{align*} &\cos \theta_1 \sin \theta_1 + \cos \theta_1 \cos \alpha_1 \sin \alpha_1 + \cdots + \cos \theta_1 \cos \alpha_1 \cdots \cos \alpha_{k - 1} \sin \alpha_{k - 1} \\ &< \sqrt{k - 1} \cos \theta_1 + \frac{1}{2} \sin(2 \theta_1). \end{align*}Changing the variable names so that $\theta_{i + 1} = \alpha_i$ for all $i \ge 1$, this becomes \begin{align*} &\cos \theta_1 \sin \theta_1 + \cos \theta_1 \cos \theta_2 \sin \theta_2 \cdots + \cos \theta_1 \cos \theta_2 + \cdots \cos \theta_k \sin \theta_k \\ &< \sqrt{k - 1} \cos \theta_1 + \frac{1}{2} \sin(2 \theta_1). \end{align*}We want to show that the LHS in the above inequality is less than $\sqrt{k}$ to complete the inductive step. Thus, to show that, it suffices to prove that for all integers $k \ge 2$, \[\sqrt{k - 1} \cos \theta + \frac{1}{2} \sin (2\theta) \le \sqrt{k}.\]We will prove a restated statement: for all positive real numbers $a$ and any $\theta \in [-\pi, \pi]$, \[(a + \sin \theta) \cos \theta \le \sqrt{a^2 + 1}.\]We may assume that $\theta \ge 0$ because if $\theta$ is negative, we can negate $\theta$ to obtain a larger expression on the left hand side. So, we may assume that both sides are positive. Squaring, \[(a + \sin \theta)^2 \cos^2 \theta \le a^2 + 1.\]Let $s = \sin \theta$. So, $s \in [0, 1]$. Letting $\cos^2\theta = 1 - s^2$ and expanding, this is equivalent to \[a^2 + 2as + s^2 - s^2a^2 - 2as^3 - s^4 \le s^2 + 1.\]One can show that this is equivalent to \[(s^2 + as - 1)^2 + s^2 \ge 0.\]Thus, our inductive step is complete.

Notice the following Cauchy-Schwarz: $(1+x_1^2+\dots+x_i^2)(n+1+\dots +1)\ge (\sqrt n + x_1 +\dots +x_i)^2$. This means our expression is at most: \[\sum_i \frac{x_i(n+i)}{(\sqrt n +x_1+\dots x_i)^2}< 2n\sum_i \frac{x_i}{(\sqrt n +x_1+\dots x_i)^2}<2n\int_{0}^{x_1+\dots+x_n}\frac1{(\sqrt n +t)^2}dt<2n\int_{\sqrt n}^\infty\frac1{t^2}dt=2\sqrt n\] The bounds used are extremely rough, so someone might be able to fix it to $\sqrt n$.