\textbf{First proof, algebraic (mine)} We proceed by induction on $n$. The base case, $n=1$ is clear since $0<\frac 13$. When $n=2,$ this is also clear because $a_2(1-a_2)\le \frac 14 < \frac 13$ Inductive step: suppose $n\ge 3$. Let $S=a_1+\cdots+a_{n-1}$, then Thus, $$\sum_{k=1}^n \frac{a_k}{1-a_k} (a_1+\cdots+a_{k-1})^2 = a_n(1-a_n) + \sum_{k=1}^{n-1} \frac{a_k}{1-a_k} (a_1+\cdots+a_{k-1})^2 $$ $$= S(1-S) + S\sum_{k=1}^{n-1} \frac{a_k}{S-Sa_k} (a_1+\cdots+a_{k-1})^2 \le S(1-S) + S\sum_{k=1}^{n-1} \frac{a_k}{S-a_k} (a_1+\cdots+a_{k-1})^2$$ Let $b_k=\frac{a_k}{S}$ for $1\le k\le n-1$. Then $b_1+\cdots+b_{n-1}=1$ Then $$ S\sum_{k=1}^{n-1} \frac{a_k}{S-a_k} (a_1+\cdots+a_{k-1})^2 = S\sum_{k=1}^{n-1} \frac{Sb_k}{S-Sb_k} S^2(b_1+\cdots+b_{k-1})^2 < \frac{S^3}{3}$$ Hence we need to show $S(1-S)<\frac{1-S^3}{3}=(1-S)\frac{1+S+S^2}{3}$ which holds since $1+S+S^2-3S=(S-1)^2$ \textbf{Second proof, combinatorial} Consider three random reals $x,y,z$ in the interval $[0,1)$. Let $$X_j=[a_1+\cdots+a_{j-1}, a_1+\cdots+a_j)$$ Then $X_1, \cdots, X_n$ partition $[0,1)$ Note $\frac 13 = Pr(x>y \text{ and } x>z)$ Also, $Pr(x>y \text{ and } x>z) = \sum\limits_{i=1}^n Pr(x\in X_i \text{ and } x>y \text{ and } x>z) = $ $$\sum\limits_{i=1}^n Pr(x\in X_i) \left(Pr(\text{ both } y,z \text{ in } X_i) + Pr(\text{ one } y,z \text{ in } X_i) + Pr(\text{ none of } y,z \text{ in } X_i)\right)$$ $$=\sum\limits_{i=1}^n \left(\frac{a_i^3}{3} + a_i^2 (a_1+\cdots+a_{i-1}) + a_i(a_1+\cdots+a_{i-1})^2\right)$$ Since if $y\in X_i, Z\in X_1\cup X_2\cup \cdots X_{i-1}$ then $Pr(x>y)=\frac 12$ and there is another case where $z\in X_i$. Note $$\frac{a_i}{1-a_i} (a_1+\cdots+a_{i-1})^2 - a_i\left(\frac{a_i^2}{3} + a_i(a_1+\cdots+a_{i-1}) + (a_1+\cdots+a_{i-1})^2\right)$$ $$=a_i^2\left( \frac{(a_1+\cdots+a_{i-1})^2}{1-a_i} - (a_1+\cdots+a_{i-1}) - \frac{a_i}{3}\right)$$ Note $a_1+\cdots+a_{i-1} < 1-a_i$, so $ \frac{(a_1+\cdots+a_{i-1})^2}{1-a_i} < a_1+\cdots+a_{i-1}$ Thus $$\sum\limits_{i=1}^n \frac{a_i}{1-a_i} (a_1+\cdots+a_{i-1})^2 < \sum\limits_{i=1}^n \left(\frac{a_i^3}{3} + a_i^2 (a_1+\cdots+a_{i-1}) + a_i(a_1+\cdots+a_{i-1})^2\right) = \frac 13$$ Remark: Both proofs may also be seen as $$\sum\limits_{k=1}^n \frac{a_k}{1-a_k} (a_1+\cdots+a_{k-1})^2 < \frac 13 (a_1+\cdots+a_n)^3$$

Note that the first term of the given sum is, by definition, $0$. Note that $\frac{a_k}{1-a_k}(a_1+\ldots+a_{k-1})^2=(a_ka_1+\ldots+a_ka_{k-1}) \cdot \frac{a_1+\ldots+a_{k-1}}{a_1+\ldots+a_{k-1}+a_{k+1}+\ldots+a_n}=$ $(a_ka_1+\ldots+a_ka_{k-1})-(a_ka_1+\ldots+a_ka_{k-1}) \cdot \frac{a_{k+1}+\ldots+a_n}{a_1+\ldots+a_{k-1}+a_{k+1}+\ldots+a_n}<$ $(a_ka_1+\ldots+a_ka_{k-1})(1-a_{k+1}-\ldots-a_n)$ for all $k \leq n-1$, and for $k=n$ the last term is equal to $a_n(1-a_n)$, because of the given constraint. Hence, summing all these we have that the given sum is less than $a_n(1-a_n)+\sum_{1 \leq i <j \leq n-1}a_ia_j-\sum_{1 \leq i<j<k \leq n}a_ia_ja_k,$ which in turn is equal to $a_n(1-a_n)+(1-a_n)\sum_{1 \leq i <j \leq n-1}a_ia_j-\sum_{1 \leq i<j<k \leq n-1}a_ia_ja_k$ Now, note that (all sums are cyclic) $3\sum_{1 \leq i<j<k \leq n-1}a_ia_ja_k=\sum a_i \sum a_ia_j-\sum a_i^2a_j=\sum a_i \sum a_ia_j-(\sum a_i \cdot \sum a_i^2-\sum a_i^3)$ and so we just need to prove that $3a_n(1-a_n)+3(1-a_n)\sum_{1 \leq i <j \leq n-1}a_ia_j-(1-a_n)\sum a_ia_j+\sum a_i \sum a_i^2-\sum a_i^3<1$ Note that the LHS is equal to $3a_n(1-a_n)+(1-a_n)(\sum a_i)^2-\sum a_i^3=3a_n(1-a_n)+(1-a_n)^3-\sum a_i^3=1-a_n^3-\sum a_i^3,$ which is clearly less than $1$, and hence we are done.

Claim. Let $x,y$ be nonnegative reals such that $x<1$ and $x+y\le 1$. Then $$\frac{xy^2}{1-x}\le\frac{1}{3}[(x+y)^3-y^3]$$with equality if and only if $x=0$. Proof. If $x=0$ we obtain equality, so for now assume $x>0$. We see that $$\frac{xy^2}{1-x}<\frac{1}{3}[(x+y)^3-y^3]$$$$\iff\frac{xy^2}{1-x}<\frac{x^3+3x^2y+3xy^2}{3}$$$$\iff\frac{xy^2}{1-x}-xy^2<\frac{x^3+3x^2y}{3}$$$$\iff\frac{x^2y^2}{1-x}<\frac{x^3+3x^2y}{3}$$$$\iff\frac{y^2}{1-x}<\frac{x+3y}{3}.$$This is true since $$\frac{y^2}{1-x}\le\frac{y^2}{y}=y<\frac{x+3y}{3}. \square$$ Now let $S_i=a_1+\ldots a_i$ for $i=0,1,\ldots,n$. Since $n>1$ and all $a_k$ are positive we have $a_k<1$ for all $k=1,\ldots,n$. It follows that each $k=1,\ldots,n$ has $$\frac{a_k}{1-a_k}(a_1+a_2+\cdots+a_{k-1})^2<\frac{1}{3}(S_k^3-S_{k-1}^3).$$Therefore $$\sum_{k=1}^n\frac{a_k}{1-a_k}(a_1+a_2+\cdots+a_{k-1})^2<\sum_{k=1}^n\frac{1}{3}(S_k^3-S_{k-1}^3)=\frac{1}{3}(S_n^3-S_0^3)=\frac{1}{3}.$$

admechii wrote: Let $n\geq 2$ be an integer and let $a_1, a_2, \ldots, a_n$ be positive real numbers with sum $1$. Prove that $$\sum_{k=1}^n \frac{a_k}{1-a_k}(a_1+a_2+\cdots+a_{k-1})^2 < \frac{1}{3}.$$ Let $s_j=a_1+a_2+\ldots+a_j$ for $j\in\{0,1,\ldots,n\}$. We have \begin{align*} 0<\frac{a_k(1-a_k)}{3}+s_{k-1}(1-s_k)=\frac{a_k(1-a_k)}{3}+s_{k-1}(1-a_k-s_{k-1})=\frac{a_k(1-a_k)}{3}+s_{k-1}(1-a_k)-s_{k-1}^2\\ 0<\frac{a_k^2(1-a_k)}{3}+s_{k-1}a_k(1-a_k)-a_ks_{k-1}^2\\ s_{k-1}^2=\frac{a_k^2(1-a_k)}{3}+s_{k-1}a_k(1-a_k)+(1-a_k)s_{k-1}^2\\ \frac{a_k}{1-a_k}s_{k-1}^2<\frac{a_k^3}{3}+s_{k-1}a_k^2+s_{k-1}^2a_k=\frac{1}{3}\left((s_{k-1}+a_k)^3-s_{k-1}^3\right)=\frac{1}{3}\left(s_k^3-s_{k-1}^3\right). \end{align*}Thus \[ \sum_{k=1}^n\frac{a_k}{1-a_k}s_{k-1}^2<\sum_{k=1}^n\frac{1}{3}\left(s_k^3-s_{k-1}^3\right)=\frac{1}{3}\left(s_n^3-s_0^3\right)=\frac{1}{3}. \]

Note: If this solution is correct, the title can be considered a spoiler. I suggest changing it.

Solution by integration. [asy][asy] size(8cm,0); draw((0,0)--(0,1.2),EndArrow(8)); draw((0,0)--(1.2,0),EndArrow(8)); pair f(real x){ return (x,x^2); } draw(graph(f,0,1.05)); real s[] = {0,0.2,0.35,0.4,0.55,0.8,0.9,1}; pair P = (s[4],s[4]^2); pair Q = (s[5],s[4]^2+(s[5]-s[4])*2*s[4]); fill((s[4],0) -- (s[5],0) -- Q -- P -- cycle, yellow); draw(P--Q, red+linewidth(1)); for(int i=1; i<s.length; ++i){ draw((s[i-1],0) -- (s[i],0) -- (s[i],s[i-1]^2) -- (s[i-1],s[i-1]^2) -- cycle); label("$a_" + (string)i + "$", ((s[i-1]+s[i])/2,0), S); } draw((s[7], s[7]^2) -- (s[7],1), dashed); label("$y=x^2$", (1.05,1.1025), NE); [/asy][/asy] Let $s_i = a_1+a_2+\dots+a_i$. Draw the rectangles of width $a_1,a_2,\dots,a_n$ as illustrated above, and for each $i$, draw a tangent at $(s_{i-1}, s_{i-1}^2)$, which meets the line $x=s_i$ at point $(s_i, s_{i-1}^2 + 2s_{i-1}a_i)$. The shaded trapezoid has area $$\frac 12 (s_{i-1}^2 + s_{i-1}^2 + 2s_{i-1}a_i)a_i = a_is_{i-1}s_i.$$Therefore, we have the inequality $$\sum_{i=1}^n a_is_{i-1}s_i < \int_0^1 x^2\ \text dx = \frac 13$$We claim that this is actually sharper than the desired inequality. It suffices to show that \begin{align*} \frac{a_is_{i-1}^2}{1-a_i} &\leq a_is_{i-1}s_i \\ \frac{s_{i-1}}{s_i} &\leq (1-a_i) \\ 1 - \frac{a_i}{s_i} &\leq 1-a_i, \end{align*}which is true because $s_i\leq 1$.

Induct on $n$. Our base case is $n = 2$, where we have $$\frac{a_2}{1-a_2}\cdot a_1^2 = \frac{a_2}{a_1} \cdot a_1^2 = a_2\cdot a_1 = a_1(1-a_1) = -\left(a_1 - \frac{1}{2}\right)^2 + \frac{1}{4} \leq \frac{1}{4} < \frac{1}{3}.$$Now, assume the result is true for $n = m$ for some $m \geq 2$. Then, note that $$\sum_{k=2}^{m+1}\frac{a_k}{1-a_k}(a_1 + \cdots + a_{k-1})^2 = \frac{a_{m+1}}{1-a_{m+1}}(a_1 + \cdots + a_m)^2 + \sum_{k=2}^{m}\frac{a_k}{1-a_k}(a_1 + \cdots + a_{k-1})^2 = \frac{a_{m+1}}{1-a_{m+1}}(1-a_{m+1})^2 + \sum_{k=2}^{m}\frac{a_k}{1-a_k}(a_1 + \cdots + a_{k-1})^2 = a_{m+1}(1-a_{m+1}) + \sum_{k=2}^{m}\frac{a_k}{1-a_k}(a_1 + \cdots + a_{k-1})^2.$$Let $c = 1 - a_{m+1}$, and define $b_i := \frac{1}{c}a_i$ for $1\leq i\leq m.$ Then, $$\sum_{k=2}^{m}\frac{a_k}{1-a_k}(a_1 + \cdots + a_{k-1})^2 = \sum_{k=2}^{m}\frac{cb_k}{1-cb_k}(cb_1 + \cdots + cb_{k-1})^2 = c^3\sum_{k=2}^{m}\frac{b_k}{1-cb_k}(b_1 + \cdots + b_{k-1})^2 < c^3\sum_{k=2}^{m}\frac{b_k}{1-b_k}(b_1 + \cdots + b_{k-1})^2,$$since $0 < c < 1$ (if $c = 1$, then $a_{m+1} = 0$, absurd, and if $c = 0$, then $a_{m+1} = 1$, so all the other $a_i$'s are $0$, absurd). However, by the inductive hypothesis, since $$\sum_{k=1}^{m}b_i = \frac{1}{1-a_{m+1}}\sum_{k=1}^{m}a_i = \frac{1}{1-a_{m+1}}(1-a_{m+1}) = 1,$$and the $b_i$ are positive real numbers since $c > 0$ as we previously discussed, we have that the sum $\sum_{k=2}^{m}\frac{b_k}{1-b_k}(b_1 + \cdots + b_{k-1})$ is less than $\frac{1}{3}$, and therefore the expression $\sum_{k=2}^{m}\frac{a_k}{1-a_k}(a_1 + \cdots + a_{k-1})^2$ is less than $\frac{1}{3}c^3 = \frac{1}{3}(1 - a_{m+1})^3.$ Thus, $$\sum_{k=2}^{m+1}\frac{a_k}{1-a_k}(a_1 + \cdots + a_{k-1})^2 < a_{m+1}(1-a_{m+1}) + \frac{1}{3}(1-a_{m+1})^3 = \frac{1}{3}(3a_{m+1} - 3a_{m+1}^2) + \frac{1}{3}(1 - 3a_{m+1} + 3a_{m+1}^2 - a_{m+1}^3) = \frac{1}{3}(1-a_{m+1})^3 < \frac{1}{3},$$as $a_{m+1} > 0$, as desired.

Let $L$ denote the LHS of the inequality. For any index $k$, let $S_k=\sum_{i=1}^{k-1} a_i$ and $T_k=\sum_{j=k+1}^n a_j$. Empty sums have value 0. Note that $S_k+T_k=1-a_k$ for all $k$. We denote the elementary symmetry polynomial of degree $m$ in the $a_k$ by $\sigma_m$. Now, by direct expansion we have $$\sum_{k=1}^n a_kS_k = \sum_{k=1}^n a_kT_k = \sum_{1 \leq i<j \leq n} a_ia_j = \sigma_2$$Thus, $$\sum_{k=1}^n \frac{a_k(S_k^2-T_k^2)}{1-a_k} = \sum_{k=1}^n a_k(S_k-T_k) = 0$$Therefore we can write $L$ as $$L=\sum_{k=1}^n \frac{a_kS_k^2}{1-a_k}=\sum_{k=1}^n \frac{a_kT_k^2}{1-a_k}$$Adding these two, we get \begin{align*} 2L &= \sum_{k=1}^n \frac{a_k(S_k^2+T_k^2)}{1-a_k} \\ &= \sum_{k=1}^n \frac{a_k(S_k^2+T_k^2)}{S_k+T_k} \\ &= \sum_{k=1}^n \frac{a_k(S_k+T_k)^2-2a_kS_kT_k}{S_k+T_k} \\ &= \sum_{k=1}^n a_k(S_k+T_k) - \sum_{k=1}^n\frac{2a_kS_kT_k}{S_k+T_k} \\ &= 2 \sigma_2 - \sum_{k=1}^n\frac{2a_kS_kT_k}{1-a_k} \\ &< 2 \sigma_2-\sum_{k=1}^n 2a_kS_kT_k \end{align*}Now by direct expansion, we see that $$\sum_{k=1}^n a_kS_kT_k = \sum_{1 \leq i <k<j \leq n} a_ia_ka_j = \sigma_3$$All in all, we get $$L<\sigma_2-\sigma_3$$Now, recall (possible from JEE times) the following identity: $$\sum_{k=1}^n a_k^3 = \sigma_1^3-3\sigma_1 \sigma_2 + 3 \sigma_3$$Since $\sigma_1=1$ here, we have $$1-\sum_{k=1}^n a_k^3 = 3(\sigma_2-\sigma_3) >3L$$Since all the $a_i$ are positive, we finally get $$L=\sum_{k=1}^n \frac{a_k}{1-a_k}(a_1+a_2+\cdots+a_{k-1})^2 < \frac{1}{3}$$as required.

Here is my solution: https://calimath.org/pdf/ISL2021-A5.pdf And I uploaded the solution with motivation to: https://youtu.be/M-FGvlrMdWg

Cali.Math wrote: Here is my solution: see And I uploaded the solution with motivation to: see here

Denote $S_i=a_1+\ldots+a_i$ for each $1 \le i \le n$, and denote $f(x)=x^2$. Note that the area of the right trapezoid with height $a_i$ and bases $S_{i-1}^2$ and $S_{i-1}^2+a_if'(S_{i-1})$ is $a_iS_{i-1}S_i$. Also, when positioned in the coordinate plane so that three of the vertices are $(S_{i-1}, 0)$, $(S_i, 0)$, and $(S_{i-1}, S_{i-1}^2)$, this trapezoid is completely under the curve $y=f(x)$. Summing over $i$, \[ \sum_{i=1}^n a_iS_{i-1}S_i < \int_0^1 x^2 \ dx = \frac{1}{3}. \]It suffices to show that $a_iS_{i-1}S_i \ge \frac{a_i}{1-a_i} S_{i-1}^2$ for all $1 \le i \le n$. We can write \[ a_i \frac{S_i}{S_{i-1}} = a_i \frac{S_{i-1}+a_i}{S_{i-1}} \ge a_i \frac{(1-a_i)+a_i}{1-a_i} = \frac{a_i}{1-a_i} \iff a_iS_{i-1}S_i \ge \frac{a_i}{1-a_i} S_{i-1}^2. \]Hence, \[ \sum_{i=1}^n \frac{a_i}{1-a_i} S_{i-1}^2 \le \sum_{i=1}^n a_iS_{i-1}S_i < \frac{1}{3}, \]as desired.

We begin by proving a lemma: Lemma: If $0\le b\le 1-a<1$ and $0<a<1$ then \[\frac{ab^2}{1-a}<\frac{(a+b)^3-b^3}{3}.\]Proof. Clearly, we have $a(a-1)<0\le 3b(1-a-b)$. Rearranging, we find that $0<-a^2-3ab-3b^3+a+3b$. We can multiply by $a^2$ and add $3ab^2$ on both sides to get \[3ab^2\le (a^3+3a^2b+3ab^3)(1-a)\]which is the result. Now, letting $s_{k}=a_1+\dots+a_k$, we see using the lemma that \begin{align*} \sum_{k=1}^n \frac{a_k}{1-a_k}(a_1+a_2+\cdots+a_{k-1})^2 &= \sum_{k=1}^n \frac{a_ks_{k-1}^2}{1-a_k}\\ &< \sum_{k=1}^n \frac{s_k^3-s_{k-1}^3}{3} \\ &= \frac{s_n^3-s_0^3}{3}\\ &= \frac{1}{3} \end{align*}as desired.

Weaker: For any $a_1,\dots,a_n\in\mathbb{R}^+$ with sum $c\in(0,1]$, \[\sum_{k=1}^n\frac{a_k}{c-a_k}(a_1+\dots+a_{k-1})^2<\frac{1}{2}c^2.\]Induct on $n$; the base case $n=2$ gives \[\frac{a_2}{c-a_2}(a_1)^2=a_2(c-a_2)\le\frac{1}{4}c^2<\frac{1}{2}c^2.\]Assume the assertion holds for $n$; we want to show that \[\sum_{k=1}^{n+1}\frac{a_k}{c-a_k}(a_1+\dots+a_{k-1})^2<\frac{1}{2}c^2.\]Let $d:=a_1+\dots+a_n$; the inductive assumption implies \[\sum_{k=1}^n\frac{a_k}{c-a_k}(a_1+\dots+a_{k-1})^2<\sum_{k=1}^n\frac{a_k}{d-a_k}(a_1+\dots+a_{k-1})^2<\frac{1}{2}d^2,\]so it suffices to show \[\frac{a_{n+1}}{c-a_{n+1}}(a_1+\dots+a_n)^2=a_{n+1}(c-a_{n+1})\le\frac{1}{2}c^2-\frac{1}{2}d^2.\]But $a_{n+1}=c-d$, turning the above into $d(c-d)\le\tfrac{1}{2}c^2-\tfrac{1}{2}d^2$―clear. Remark. Discovered by attempting Force-Overlaid InductionTM with $\tfrac{1}{2}c^2$ replaced by $f(c)$ to give $\tfrac{f(c)-f(d)}{c-d}\ge d$, where $d\to c^-$ gives $f'(c)\ge c$ or $f(c)\ge\tfrac{1}{2}c^2+C$. From a calculus perspective, $f(c)=\tfrac{1}{2}c^2$ is sufficient because $\tfrac{f(c)-f(d)}{c-d}=f'(x)=x\ge d$ for some $x\in[c,d]$, by MVT. Sadly, such a function with $f(1)\le\tfrac{1}{3}$ doesn't appear to exist.

Here is a very interesting solution by drawing a graph:

2021 A5 wrote: Let $n\geq 2$ be an integer and let $a_1, a_2, \ldots, a_n$ be positive real numbers with sum $1$. Prove that$$\sum_{k=1}^n \frac{a_k}{1-a_k}(a_1+a_2+\cdots+a_{k-1})^2 < \frac{1}{3}.$$ We homegenize to prove the inequality $\sum_{k=1}^{n} \frac{a_k}{1-a_k}\left(\sum_{i=1}^{k-1} a_i\right)^2 < \frac{1}{3}\left(\sum_{i=1}^n a_i \right)^3$ Lemma: For $x \leq 1$ and $x+y \leq 1$ we have the following inequality,

Now by using the lemma, the sum telescopes and we get the desired bound.

We do induction. Let the last element be $x$. Observe that $\frac{a(1-x)}{1-a-x}-\frac{a}{1-a} = \frac{a^2x}{(1-a)(1-a-x)}\ge 0$. Hence by induction the first $n-1$ summans are at most $(1-x) \frac{(1-x)^2}{3}$. Hence it is enough to prove $x(1-x) + \frac{(1-x)^3}{3} = \frac{1}{3} - \frac{x^3}{3}\leq \frac{1}{3}$