# Maximum value of $\sqrt{1-\sqrt{a_1}} + \sqrt{1-\sqrt{a_2}} + \cdots + \sqrt{1-\sqrt{a_n}}$ for $a_i \in [0,1], a_1+a_2+ \cdots +a_n = 2$

Given $$n \ge 3$$ real numbers $$a_1, a_2, \cdots, a_n \in [0,1]$$ such that $$a_1 + a_2+ \cdots +a_n =2$$, find the maximum value of: $$P =\sqrt{1-\sqrt{a_1}} + \sqrt{1-\sqrt{a_2}} + \cdots + \sqrt{1-\sqrt{a_n}}.$$

So far my only guess is that the equality occurs when there are $$n-x$$ numbers equal to $$0$$ and $$x$$ numbers equal to $$\dfrac{2}{x}$$, then we have: $$P \le n-x + x \sqrt{1-\sqrt{\frac{2}{x}}} = n-x + \sqrt{x^2-x\sqrt{2x}}$$ The function above reaches maximum at $$x=3$$, so we have: $$P \le n-3 + \sqrt{9 -3\sqrt{6}}$$ And I didn't know how to proceed any further.

• For whatever it's worth, that's a good guess. If tries zillions of combinations of a1, a2,...a6, then you'll see the combinations resulting in near the maximum with 3 zeros and 3 values close to 2/3. (Take a large sample from a Dirichlet distribution, multiply by 2, and then filter out the samples where the maximum is greater than one.)
– JimB
Feb 8, 2022 at 5:08

A brute force method: $$f(x)=\sqrt{1-\sqrt x}$$, then, $$f'(x)=-\frac{1}{4\sqrt{x}\sqrt{1-\sqrt{x}}}$$, $$f''(x)=\frac{2-3\sqrt x}{16\sqrt{(1-\sqrt x)x}^3}$$. Therefore, the function $$f(x)$$ is convex for $$x<4/9$$ and concave for $$x>4/9$$. So, we can make $$f(x_1)+f(x_2)$$ (WLOG $$x_1\le x_2$$) larger we do the adjustment:

(a) If $$0\le x_1\le x_2\le 4/9$$, $$f(x_1)+f(x_2)\le f(x_1-\delta)+f(x_2+\delta)$$ for all $$\delta\in[0,\min(4/9-x_2,x_1)]$$.

(a) If $$4/9\le x_1\le x_2\le 1$$, $$f(x_1)+f(x_2)\le f(x_1+\delta)+f(x_2-\delta)$$ for all $$\delta\in[0,\frac{x_2-x_1}2]$$.

Therefore, $$\sum_{i}f(x_i)$$ reach maximum when it can't be adjusted by the method above. Specifically:

(a) all the $$x\ge 4/9$$'s are equal.

(b) all the other $$x\le 4/9$$ are hitting the boundary (that is, being $$0$$ or one single value.)

So, we have $$0=x_1=\dots=x_k\le x_{k+1}<4/9\le x_{k+2}=\dots=x_n$$.

Since all $$x_i\le 1$$, we have $$2\le n-k\le 4$$. We have three cases:

Case 1: $$n-k=2$$. So, we have $$0=x_1=\dots=x_{n-3}\le x_{n-2}<4/9\le x_{n-1}=x_n$$. Let $$x_n=x$$, and we have $$x_n\ge 7/9$$. Therefore, we have

\begin{align}&\sum_i f(x_i)=n-3+\sqrt{1-\sqrt{x_{n-2}}}+\sqrt{1-\sqrt{x_{n-1}}}+\sqrt{1-\sqrt{x_{n}}}\\ \le &n-3+1+2\sqrt{1-\sqrt{7/9}}\le n-3+1.236\dots\end{align}

Notice that $$n-3+\sqrt{9-3\sqrt{6}}=n-3+1.285\dots$$, so this case is smaller.

Case 2: $$n-k=3$$. So, we have $$0=x_1=\dots=x_{n-4}\le x_{n-3}<4/9\le x_{n-2}= x_{n-1}=x_n$$. Let $$x_n=x$$, and we have $$14/27\le x_n\le 2/3$$. Therefore, we have

\begin{align}&\sum_i f(x_i)=n-4+\sqrt{1-\sqrt{2-3x}}+3\sqrt{1-\sqrt{x}}\end{align}

Let $$g(x)=\sqrt{1-\sqrt{2-3x}}+3\sqrt{1-\sqrt{x}}$$. Therefore, $$g'(x)=\frac{3}{4}(\frac 1{\sqrt{2-3x}\sqrt{1-\sqrt{2-3x}}}-\frac 1{\sqrt{1-x}\sqrt{1-\sqrt{x}}})$$ Notice that $$\sqrt{x}\sqrt{1-\sqrt{x}}$$ is an increasing function in $$[0,1]$$. Also, $$2-3x\le x$$ since $$14/27\le x_n\le 2/3$$. Therefore, we have $${\sqrt{2-3x}\sqrt{1-\sqrt{2-3x}}}\le {\sqrt{1-x}\sqrt{1-\sqrt{x}}}$$, and thus $$\frac 1{\sqrt{2-3x}\sqrt{1-\sqrt{2-3x}}}-\frac 1{\sqrt{1-x}\sqrt{1-\sqrt{x}}}\ge 0$$, $$g'(x)\ge 0$$. Therefore, in this case, $$x=2/3$$ get the maximum value, which is your proposed solution \begin{align}&\sum_i f(x_i)=n-4+\sqrt{1-\sqrt{2-3\times 2/3}}+3\sqrt{1-\sqrt{2/3}}=n-3+\sqrt{9-6\sqrt 3}\end{align}

Case 3: $$n-k=4$$. So, we have $$0=x_1=\dots=x_{n-5}\le x_{n-4}<4/9\le x_{n-3}=\dots =x_n$$. Let $$x_n=x$$, and we have $$4/9\le x_n\le 1/2$$. Therefore, we have

\begin{align}&\sum_i f(x_i)=n-5+\sqrt{1-\sqrt{2-4x}}+4\sqrt{1-\sqrt{x}}\end{align}

Let $$g(x)=\sqrt{1-\sqrt{2-4x}}+4\sqrt{1-\sqrt{x}}$$. Therefore, $$g'(x)=\frac 1{\sqrt{2-4x}\sqrt{1-\sqrt{2-4x}}}-\frac 1{\sqrt{1-x}\sqrt{1-\sqrt{x}}}$$. Similar to the previous case, $$g'(x)\ge 0$$, and we take $$x=1/2$$. Therefore, we have

\begin{align}&\sum_i f(x_i)=n-5+1+4\sqrt{1-\sqrt{1/2}}=n-3+1.164...\end{align} which is also smaller.

Therefore, the maximum value is $$n-3+\sqrt{9-3\sqrt 6}$$

• You have a typo your wrote $2\leq n-k\geq 4$ Feb 8, 2022 at 9:01
• @ErikSatie Thank you for pointing out. Feb 8, 2022 at 13:17
• you have (perhaps?) another typo at the beginning where $\min(9/4\cdots)$ Feb 8, 2022 at 15:02
• @ErikSatie Thank you for pointing out again! Feb 8, 2022 at 18:11

Hint : use the EV method with :

$$f(x)=\sqrt{1-\sqrt{1-\frac{x}{x+1}}}$$

We have on $$(0,\infty)$$:

$$f'''(x)>0$$

With the constraint (for $$0):

$$\sum_{i=1}^{n}\frac{1-x_i}{x_i}=2\,,\sum_{i=1}^{n}\left(\frac{1-x_i}{x_i}\right)^2=\operatorname{constant}$$

Now use corollary 1.4 (see http://emis.maths.adelaide.edu.au/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf )

• EV theorem states that the maximum/minimum occurs when $n - 1$ variables are equal. However, in the OP, the minimum occurs when $n-3$ variables are zero and the remaining $3$ variables are equal to $2/3$. How did you apply EV theorem? Feb 11, 2022 at 2:54
• @RiverLi I think it works for small case like $n=3$ no ? Feb 11, 2022 at 8:53
• You may delete it if it only works for $n=3$. You do not provide a solution for general $n$ (like the other answer). In the future, you find a solution, and you can reopen this answer. Feb 11, 2022 at 9:28