Given scalars $p_1\geq p_2\geq \cdots \geq p_r > 0$, can we find a solution for following problem?
\begin{align} \text{minimize} & & & \sum_{j=1}^{r} p_j (1-t_j)^2 \\ \text{s.t.} \\ & & & \forall j:t_j\geq 0 \\ & & & \sum_{j=1}^{r} t_j = 1 \end{align}
- If finding a solution is impossible in general, which lower bounds we can derive for the objective function?
Hint
The function achieves maximum only when the variables $t_i \in \{0, 1\}$ (why?), so $t_r=1$.
Added: For the minimum, easiest way seems Cauchy-Schwarz inequality: $$\left(\sum_{j=1}^rp_j (1-t_j)^2\right)\left( \sum_{j=1}^r \frac1{p_j} \right) \ge \left( r-1\right)^2$$
By the equality condition of CS inequality, for the minimum we need all $\dfrac{p_j(1-t_j)^2}{1/p_j} = p_j^2(1-t_j)^2$ to be the same. i.e. $p_j(1-t_j) = k$ or $t_j = 1-\dfrac{k}{p_j}$. We can solve for $k$ using the condition $\sum t_j = 1$, whence $k = \dfrac{r-1}{\sum 1/p_j}$. Putting all that together, we have $$\sum p_j (1-t_j)^2 \ge \frac{(r-1)^2}{\sum 1/p_j}$$ with equality iff $t_j = 1-\dfrac{r-1}{p_j \sum 1/p_j}$.