\begin{align*} a(1-a^k)+b(1-(a+b)^k)+c(1-(a+b+c)^k) &< \int_{0}^{a} (1-x^k) dx + \int_{a}^{a+b} (1-x^k) dx +\int_{a+b}^{a+b+c} (1-x^k) dx \\ &= \int_{0}^{a+b+c} (1-x^k) dx \\ &\le \int_{0}^{1} (1-x^k) dx = \frac{k}{k+1} \end{align*} Remark. 2018 Serbia TST @below It's easy but pointless to rephrase with "simple" method.

oVlad wrote: Let $k$ be a positive integer, and let $a,b$ and $c$ be positive real numbers. Show that \[a(1-a^k)+b(1-(a+b)^k)+c(1-(a+b+c)^k)<\frac{k}{k+1}.\] * * * Simple proof?

The inequality is equivalent to \[ a^{k+1} + b(a+b)^{k} + c(a+b+c)^k + \frac{k}{k+1} > a + b + c \]We have the following lemma Let $a,b > 0$ and $k \in \textrm{N}$ then \[ a^{k+1} + b(a+b)^k \ge \frac{(a+b)^{k+1}}{k+1} \] To prove this lemma, we can write the inequality as \[ \left(\frac{a}{a+b}\right)^{k+1} + \frac{b}{a+b} \ge \frac{1}{k+1} \]Let $\displaystyle x = \frac{a}{a+b}$ then the above inequality becomes \[ x^{k+1} + 1 - x \ge \frac{1}{k + 1} \Longleftrightarrow x^{k+1} + \frac{k}{k+1} \ge x \]However, this is true as it is the direct application of AM-GM inequality for $(k+1)$ positive numbers as follows \[ x^{k+1} + \underbrace{\frac{1}{k+1} + \ldots + \frac{1}{k+1}}_\text{$k$ terms} \ge (k+1)\cdot \left(\frac{x^{k+1}}{(k+1)^{k+1}}\right)^{\frac{1}{k+1}} = x \]This completes our proof for the lemma. Back to the original question, using the above lemma, one has \[ a^{k+1} + b(a+b)^{k} + c(a+b+c)^k \ge \frac{(a+b)^{k+1}}{k+1} + c(a+b+c)^k \]We will prove the stronger inequality \[ \frac{(a+b)^{k+1}}{k+1} + c(a+b+c)^k \ge \frac{(a+b+c)^{k+1}}{k+1} \]Let $x = \displaystyle \frac{a+b}{a+b+c}$, the inequality becomes \[ x^{k+1} + k \ge (k+1)x \]This is again a direct application of AM-GM and the equality does not occur since $x \in (0,1)$. Therefore, \[ \frac{(a+b)^{k+1}}{k+1} + c(a+b+c)^k > \frac{(a+b+c)^{k+1}}{k+1} \]Using AM-GM again, we have \[ \frac{(a+b)^{k+1}}{k+1} + c(a+b+c)^k + \frac{k}{k+1} > \frac{(a+b+c)^{k+1}}{k+1} + \frac{k}{k + 1} \ge a + b + c \]This completes our proof. We can generalise the inequality into $n$ variables as follows: for $a_i > 0$ for $i \in \{1, 2, \ldots, n\}$ and $k \in \textrm{N}$, we have \[ \sum_{i=1}^n a_i\left[1-\left(\sum_{j = 1}^i a_j\right)^k\right] < \frac{k}{k+1} \]

Rickyminer wrote: \begin{align*} a(1-a^k)+b(1-(a+b)^k)+c(1-(a+b+c)^k) &< \int_{0}^{a} (1-x^k) dx + \int_{a}^{b} (1-x^k) dx +\int_{a+b}^{a+b+c} (1-x^k) dx \\ &= \int_{0}^{a+b+c} (1-x^k) dx \\ &\le \int_{0}^{1} (1-x^k) dx = \frac{k}{k+1} \end{align*} Remark. 2018 Serbia TST @below It's easy but pointless to rephrase with "simple" method. Nice solution, although you do have to keep in mind that this is a junior problem.

Let $a_1,a_2, \ldots ,a_n > 0$ and $k \in \textrm{N}.$ Prove that \[ \sum_{i=1}^n a_i\left[1-\left(\sum_{j = 1}^i a_j\right)^k\right] < \frac{k}{k+1} \]Let $n$ be a fixed positive integer and let $x_1,\ldots,x_n$ be positive real numbers. Prove that $$\sum_{i=1}^n a_i\left[1-\left(\sum_{j = 1}^i a_j\right)^2\right] <\frac{2}{3}.$$Serbia 2018

Rickyminer wrote: \begin{align*} a(1-a^k)+b(1-(a+b)^k)+c(1-(a+b+c)^k) &< \int_{0}^{a} (1-x^k) dx + \int_{a}^{b} (1-x^k) dx +\int_{a+b}^{a+b+c} (1-x^k) dx \\ &= \int_{0}^{a+b+c} (1-x^k) dx \\ &\le \int_{0}^{1} (1-x^k) dx = \frac{k}{k+1} \end{align*} Remark. 2018 Serbia TST @below It's easy but pointless to rephrase with "simple" method. I think there is a typo in this solution. Rickyminer wrote: \begin{align*} a(1-a^k)+b(1-(a+b)^k)+c(1-(a+b+c)^k) &< \int_{0}^{a} (1-x^k) dx + \int_{a}^{b} (1-x^k) dx +\int_{a+b}^{a+b+c} (1-x^k) dx\end{align*} It should be : \begin{align*} a(1-a^k)+b(1-(a+b)^k)+c(1-(a+b+c)^k) &< \int_{0}^{a} (1-x^k) dx + \int_{a}^{a+b} (1-x^k) dx +\int_{a+b}^{a+b+c} (1-x^k) dx\end{align*}

vietmath wrote: The inequality is equivalent to \[ a^{k+1} + b(a+b)^{k} + c(a+b+c)^k + \frac{k}{k+1} > a + b + c \]We have the following lemma Let $a,b > 0$ and $k \in \textrm{N}$ then \[ a^{k+1} + b(a+b)^k \ge \frac{(a+b)^{k+1}}{k+1} \] To prove this lemma, we can write the inequality as \[ \left(\frac{a}{a+b}\right)^{k+1} + \frac{b}{a+b} \ge \frac{1}{k+1} \]Let $\displaystyle x = \frac{a}{a+b}$ then the above inequality becomes \[ x^{k+1} + 1 - x \ge \frac{1}{k + 1} \Longleftrightarrow x^{k+1} + \frac{k}{k+1} \ge x \]However, this is true as it is the direct application of AM-GM inequality for $(k+1)$ positive numbers as follows \[ x^{k+1} + \underbrace{\frac{1}{k+1} + \ldots + \frac{1}{k+1}}_\text{$k$ terms} \ge (k+1)\cdot \left(\frac{x^{k+1}}{(k+1)^{k+1}}\right)^{\frac{1}{k+1}} = x \]This completes our proof for the lemma. Back to the original question, using the above lemma, one has \[ a^{k+1} + b(a+b)^{k} + c(a+b+c)^k \ge \frac{(a+b)^{k+1}}{k+1} + c(a+b+c)^k \]We will prove the stronger inequality \[ \frac{(a+b)^{k+1}}{k+1} + c(a+b+c)^k \ge \frac{(a+b+c)^{k+1}}{k+1} \]Let $x = \displaystyle \frac{a+b}{a+b+c}$, the inequality becomes \[ x^{k+1} + k \ge (k+1)x \]This is again a direct application of AM-GM and the equality does not occur since $x \in (0,1)$. Therefore, \[ \frac{(a+b)^{k+1}}{k+1} + c(a+b+c)^k > \frac{(a+b+c)^{k+1}}{k+1} \]Using AM-GM again, we have \[ \frac{(a+b)^{k+1}}{k+1} + c(a+b+c)^k + \frac{k}{k+1} > \frac{(a+b+c)^{k+1}}{k+1} + \frac{k}{k + 1} \ge a + b + c \]This completes our proof. We can generalise the inequality into $n$ variables as follows: for $a_i > 0$ for $i \in \{1, 2, \ldots, n\}$ and $k \in \textrm{N}$, we have \[ \sum_{i=1}^n a_i\left[1-\left(\sum_{j = 1}^i a_j\right)^k\right] < \frac{k}{k+1} \] \[ x^{k+1} + \underbrace{\frac{1}{k+1} + \ldots + \frac{1}{k+1}}_\text{$k$ terms} \ge (k+1)\cdot \left(\frac{x^{k+1}}{(k+1)^{k+1}}\right)^{\frac{1}{k+1}} ? \] We have $$x^{k+1} + \frac{k}{k+1} \ge (k+1) \sqrt[k+1]{ \frac{x^{k+1}}{(k+1)^k} }=x\sqrt[k+1]{ 1+k}>x $$