Let $n \geq 3$ be an integer. Let $a_1,a_2,\dots, a_n\in[0,1]$ satisfy $a_1 + a_2 + \cdots + a_n = 2$. Prove that $$\sqrt{1-\sqrt{a_1}}+\sqrt{1-\sqrt{a_2}}+\cdots+\sqrt{1-\sqrt{a_n}}\leq n-3+\sqrt{9-3\sqrt{6}}.$$
Problem
Source: 2018 Thailand TST 6.3
Tags: inequalities
21.04.2022 04:23
Let $a,b,c \in[0,1]$and $a+b+c= 2$. Prove that $$\sqrt{1-\sqrt{a}}+\sqrt{1-\sqrt{b}}+\sqrt{1-\sqrt{c}}\leq \sqrt{9-3\sqrt{6}}$$ Let $a,b,c\leq 1$ and $a^2 +b^2 +c^2= 2$. Prove that $$\sqrt{1-a}+\sqrt{1-b}+\sqrt{1-c} \leq \sqrt{9+3\sqrt{6}}$$
13.06.2022 04:16
Bump$\phantom{8char bruh}$
13.06.2022 04:31
https://www.infinitydots.org/testarchive/sp_tha_tst_2018_sol_en.pdf#:~:text=x1.2%20Thailand%20TST%202018%20Information%20Thailand%20TST%20is,thefollowing%20dates.%20%0FDay%201%20%3A%2025%20December%202017
13.06.2022 15:12
Very good content
23.06.2022 21:43
Let $f(x)=\sqrt{1-\sqrt{x}}$ We have $$f'(x)= -\frac 14 (x-x^{\frac 32})^{-\frac 12}$$ $$f''(x)= \frac 18 (x-x^{\frac 32})^{-\frac 32} \left(1-\frac 32 \sqrt{x}\right)$$ Thus, $f$ is convex in $[0,\frac 49]$ and concave in $[\frac 49, 1]$ We use the following smoothing algorithm. Let $A$ be the multi-set of $a_i$'s in the interval $(0,\frac 49)$ and $B$ be the multiset of $a_i$'s in the interval $[\frac 49,1]$. Our process makes sure $|A|\le 1$. If $a_i, a_j\in A$ for some $i\ne j$ we have two cases: \begin{itemize} \item if $a_i+a_j<\frac 49$, then I replace $a_i, a_j$ with $0, a_i+a_j$. This increases the desired sum because $f(a_i)+f(a_j)< f(0)+f(a_i+a_j)$. \item Otherwise, I replace them with $\frac 49, a_i+a_j-\frac 49$. This increases the sum because $f(a_i)+f(a_j) < f\left(\frac 49\right) + f\left( a_i+a_j-\frac 49\right)$ \end{itemize} This process reduces $|A|$ by 1 as long as $|A|\ge 2$. By concavity, all elements in $B$ are equal. Therefore, this boils down to maximizing $$g(S)=f(S)+kf\left(\frac{2-S}{k}\right)$$ For $\frac{2-S}{k} \in [\frac 49,1]$ and $S\in [0,\frac 49)$ We have \begin{align*} g'(S) & = f'(S)+kf'\left(\frac{2-S}{k} \right)\\ & = -\frac 14 (S-S^{\frac 32})^{-\frac 12} + k \frac{df\left( \frac{2-S}{k}\right)}{d\frac{2-S}{k}} \cdot \frac{d\frac{2-S}{k}}{dS}\\ & = -\frac 14 \left( (S-S^{\frac 32})^{-\frac 12} - \left(\frac{2-S}{k} - \left(\frac{2-S}{k}\right)^{\frac 32} \right)^{-\frac 12} \right) \end{align*} To maximize this expression, we only need to consider $S=0$ and all points $S$ such that $g'(S)=0$. Fix $k$ and consider the root to $g'(S)=0$. This $S$ also satisfy $$h(S)=S-S^{\frac 32} - \left(\frac{2-S}{k}\right) + \left(\frac{2-S}{k}\right)^{\frac 32}$$ Then $h''(S)=-\frac 34 S^{-\frac 12} + k^{-\frac 32} \frac 34 (2-S)^{-\frac 12}$. We can check $h''(S)<0$ for all $0<S<\frac 49$ and $k\ge 2$. This means there exists at most one $S$ such that $g'(S)=0$ For $k=2$, note $g'(0)=g'(\frac 23)=0$. Therefore, we can check there is no solution to $g'(S)=0$. For $k=3$, we can check $h(\frac 13) < 0 < h(\frac{7}{20})$, and we can check $g''(S)>0$ for all $\frac 13 < S < \frac{7}{20}$ For $k=4$, we can check that $g'(\frac 25)=0$. Note $k\ge 5$ is impossible because $\frac 49 \cdot 5>2$. Therefore we can check $S=0$ and finitely many points to see the inequality is true.
25.06.2022 07:54
Very effective