Maximum of $(1-q_1)(1-q_2)\ldots(1-q_n)$

I'm trying to find the maximum of $(1-q_1)(1-q_2)\ldots(1-q_n)$ where $n\ge 2$, on a the set $\{(q_1,\ldots , q_n) :q_1^2+q_2^2+\ldots+q_n^2=1 \ q_i\ge 0 \}$ (With the condition $q_i\ge0$ this is just the upper half of the sphere). This appeared to be a simple Lagrange multipliers question but calculating the derivative is becoming a problem. How is this done?

I was thinking of trying the fact that the product is the volume of a box having $(1,1\ldots 1)$ and $(q_1,\ldots ,q_n)$ at its diagonal but I can't figure out how to continue.(I'm guessing this reduces the problem to something geometrical)

Updates: Empirical evidence suggests that the maximum is when the $q_i$ are equal as pointed out by @Sabyasachi , I let WolframAlpha solve the Laplace multiplier equations for $n=2,3,4$ and got the same result(that they are equal). One solution which does not turn up in WA's solutions is that $q_3=0$ and $q_1=q_2=\frac {1} {\sqrt{2}}$ which is on the sphere and returns a larger value.

Here is a complete answer. The computations are rather long but every step is natural. Perhaps someone else can simplify the computational part of the proof.

We will show that the maximum is $M=\frac{3}{2}-\sqrt{2}$ independently of $n$, just as claimed in Macavity's comments. Let $\phi(x)=1-\sqrt{x}$ for $x\in [0,1]$. The inequality to be shown can then be restated as :

$$ \phi(x_1)\phi(x_2)\ldots \phi(x_n)\leq M, \ \text{whenever} \ x_i\geq 0, \ x_1+x_2+x_3+\ldots +x_n=1. \tag{1} $$

To show such an inequality, one of the first ideas that comes to mind is to try to show that $\phi$ satisfies properties like $\phi(x)\phi(y) \leq \phi(x+y)$ or $\phi(x+y)\leq (\phi(\frac{x+y}{2}))^2$. Unfortunately, both those are false (take $x=\frac{1}{6},y=\frac{1}{3}$). We must use a slightly corrected version of $\phi$ : let

$$ \psi(t)=\left\lbrace\begin{array}{lcl} \phi(t), & \rm{if} & t\leq \alpha, \\ \phi\big(\frac{t}{2}\big)^2, & \rm{if} & t\geq \alpha, \\ \end{array}\right.\tag{2} $$

where $\alpha=4(3-2\sqrt{2})$ is the unique solution of $\phi(t)=\phi(\frac{t}{2})^2$ in $(0,1)$.

Lemma 1. $\phi \leq \psi$ on $[0,1]$.

Lemma 2. $\psi$ satisfies $\psi(x)\psi(y) \leq \psi(x+y)$, for any $x,y\in [0,1]$ with $x+y \leq 1$.

Lemmas 1 and 2 yield $$\phi(x_1)\phi(x_2)\ldots \phi(x_n) \leq \psi(x_1)\psi(x_2)\ldots \psi(x_n) \leq \psi(x_1+x_2+x_3+\ldots +x_n) =\psi(1)=M \tag{3}$$

as wished. It will therefore suffice to show those two lemmas. This we do below.

Proof of lemma 1. We must show $\phi(t) \leq \psi(t)$ for $t\in [0,1]$. Clearly, we may assume that $t\geq\alpha$. But then

$$ \psi(t)-\phi(t)=\frac{\sqrt{t}}{2}(\sqrt{t}-\sqrt{\alpha}) \geq 0. $$

Proof of lemma 2. We must show $\psi(x)\psi(y) \leq \psi(x+y)$ for $x,y\in [0,1]$.

First case : $x+y\leq \alpha$.

We then have to show that $$ (1-\sqrt{x})(1-\sqrt{y})\leq 1-\sqrt{x+y} \tag{4} $$

Putting $a=\sqrt{x},b=\sqrt{y}$, this is equivalent to

$$ \begin{array}{cl} & (1-a)(1-b)\leq 1-\sqrt{a^2+b^2} \\ \Leftrightarrow & ab+\sqrt{a^2+b^2} \leq a+b \\ \Leftrightarrow & a^2b^2+a^2+b^2+2ab\sqrt{a^2+b^2} \leq a^2+b^2+2ab \\ \Leftrightarrow & a^2b^2+2ab\sqrt{a^2+b^2} \leq 2ab \\ \Leftrightarrow & ab+2\sqrt{a^2+b^2} \leq 2 \\ \end{array} $$

Now, from the hypotheses we have $a^2+b^2 \leq \alpha$, so

$$ ab+2\sqrt{a^2+b^2} \leq \frac{a^2+b^2}{2}+2\sqrt{a^2+b^2} \leq \frac{\alpha}{2}+2\sqrt{\alpha}=6-4\sqrt{2}+4\sqrt{2}-4=2 \tag{5} $$

Second case : $x\leq \alpha, y\leq \alpha, x+y > \alpha$.

We then have to show that $$ (1-\sqrt{x})(1-\sqrt{y})\leq \bigg(1-\sqrt{\frac{x+y}{2}}\bigg)^2 \tag{6} $$

Putting $a=\sqrt{x},b=\sqrt{y}$, this is equivalent to

$$ \begin{array}{cl} & (1-a)(1-b)\leq \bigg(1-\frac{\sqrt{a^2+b^2}}{2}\bigg)^2 \\ \Leftrightarrow & \sqrt{2(a^2+b^2)} \leq a+b+\frac{a^2+b^2}{2}-ab \\ & \\ \Leftrightarrow & 2(a^2+b^2) \leq \frac{a^4+b^4}{4}-(a^3b+ab^3)-(a^2b+ab^2)+(a^3+b^3)+ \frac{3a^2b^2}{2} +a^2+b^2-2ab \\ & \\ \Leftrightarrow & 0 \leq \frac{a^4+b^4}{4}-(a^3b+ab^3)-(a^2b+ab^2)+(a^3+b^3)+ \frac{3a^2b^2}{2} -a^2-b^2-2ab \\ \Leftrightarrow & 0 \leq \frac{(b-a)^2}{4}(a^2+b^2+4(a+b)-2ab-4) \\ \Leftrightarrow & 0 \leq a^2+b^2+4(a+b)-2ab-4 \\ \Leftarrow & 0 \leq \alpha+4(a+b)-2ab-4 \\ \Leftrightarrow & 0 \leq 4(a+b)-2ab-4\sqrt{\alpha} \ ({\rm since} \ \alpha-4=-4\sqrt{\alpha})\\ \Leftrightarrow & 2(\sqrt{\alpha}-a) \leq b(2-a) \\ \Leftrightarrow & 4(\sqrt{\alpha}-a)^2 \leq b^2(2-a)^2 \\ \Leftarrow & 4(\sqrt{\alpha}-a)^2 \leq (\alpha-a^2)(2-a)^2 \\ \Leftrightarrow & 4(\sqrt{\alpha}-a) \leq (\sqrt{\alpha}+a)(2-a)^2 \\ \end{array} $$

The last inequality is true because

$$ (\sqrt{\alpha}+a)(2-a)^2-4(\sqrt{\alpha}-a)=a(16(1-\sqrt{\alpha})+(\sqrt{\alpha}-a)(\alpha+2\sqrt{\alpha}-a)) \tag{7} $$

\newpage

Third case : One of $x,y$ is smaller than $\alpha$, the other is larger.

We can assume $x \leq \alpha \leq y \leq 1-x$. Notice that $x\leq 1-\alpha$. We then have to show that $$ (1-\sqrt{x})\bigg(1-\sqrt{\frac{y}{2}}\bigg)^2\leq \bigg(1-\sqrt{\frac{x+y}{2}}\bigg)^2 \tag{8} $$

To this end, let us put $F_1(y)=\frac{1-\sqrt{\frac{x+y}{2}}}{1-\sqrt{\frac{y}{2}}}$ for $y\in [\alpha,1-x]$. A little computation shows that

$$ {F'}_1(y)=\frac{F_2(y)}{\sqrt{y(x+y)}\bigg(1-\sqrt{\frac{y}{2}}\bigg)^2}, \ F_2(y)=\sqrt{x+y}-\sqrt{y}-\frac{x}{\sqrt{2}} \tag{9} $$

Note that ${F'}_2(y)=\frac{1}{2\sqrt{x+y}}-\frac{1}{2\sqrt{y}} \leq 0$ so $F_2$ is decreasing, and hence $F_2(y)\leq F_2(\alpha)=F_3(x)$ where $F_3(x)=\sqrt{x+\alpha}-\frac{x}{\sqrt{2}}-\sqrt{\alpha}$. Note that ${F'}_3(y)=\frac{1}{2\sqrt{x+\alpha}}-\frac{1}{\sqrt{2}} \leq 0$ and $2\sqrt{x+\alpha} \geq 2\sqrt{\alpha} > 1.5 > \sqrt{2}$, so $F_3$ is decreasing, and hence $F_3(x) \leq F_3(0)=0$. We see now that $F_2(y) \leq 0$, so $F_1$ is decreasing, and hence $F_1(y)\geq F_1(1-x)$. So in the proof of (8), we can assume that $y=1-x$ : all we need to show is

$$ (1-\sqrt{x})\bigg(1-\sqrt{\frac{1-x}{2}}\bigg)^2\leq \bigg(1-\sqrt{\frac{1}{2}}\bigg)^2 \tag{10} $$

Putting $a=\sqrt{x}$, this is equivalent to

$$ \begin{array}{cl} & (1-a)\bigg(1-\sqrt{\frac{1-a^2}{2}}\bigg)^2\leq \frac{3}{2}-\sqrt{2} \\ \Leftrightarrow &(1-a)\bigg(1-\sqrt{2(1-a^2)}+\frac{1-a^2}{2}\bigg)\leq \frac{3}{2}-\sqrt{2} \\ & \\ \Leftrightarrow & (1-a)\bigg(1+\frac{1-a^2}{2}\bigg)-(\frac{3}{2}-\sqrt{2})\leq (1-a)\sqrt{2(1-a^2)} \\ \Leftrightarrow & \frac{a^3-a^2-3a}{2}+\sqrt{2}\leq (1-a)\sqrt{2(1-a^2)} \\ \Leftarrow & \bigg(\frac{a^3-a^2-3a}{2}-\sqrt{2}\bigg)^2\leq (1-a)^22(1-a^2) \\ \end{array} $$

The last inequality is true because $a\leq\sqrt{1-\alpha} \leq 0.6=\frac{3}{5}$ and

$$ \begin{array}{cl} & 2(1-a)^2(1-a^2)-\bigg(\frac{a^3-a^2-3a}{2}-\sqrt{2}\bigg)^2 \\ =& \frac{a}{4}(-a^5+2a^4 - 3a^3 + (10-4\sqrt{2})a^2 + (4\sqrt{2} - 9)a + (12\sqrt{2}- 16)) \\ \geq& \frac{a}{4}(-a^5+2a^4 - 3a^3 +\frac{108}{25}a^2 + -\frac{84}{25}a + \frac{24}{25} \end{array} \tag{11} $$

Fourth case : Both $x$ and $y$ are larger than $\alpha$.

We then have to show that $$ \bigg(1-\sqrt{\frac{x}{2}}\bigg)^2\bigg(1-\sqrt{\frac{y}{2}}\bigg)^2\leq \bigg(1-\sqrt{\frac{x+y}{2}}\bigg)^2 \tag{12} $$

But this is simply inequality (4) of the first case, used with $(\frac{x}{2},\frac{y}{2})$ in place of $(x,y)$. And this is all OK since $\frac{x}{2}$ and $\frac{y}{2}$ are both $\leq \frac{1}{2} \leq \alpha$, so we only need to reuse our already treated first case. This concludes the proof.

${\bf 1\ }$ We begin with the following two-dimensional problem: Maximize $$f(q_1,q_2):=(1-q_1)(1-q_2)$$ under the constraints $$q_1^2+q_2^2=r^2, \quad q_1\geq0, \ q_2\geq0\ .$$ Here $r$ is a parameter, $0<r\leq1$. Analyzing the graphs of the functions $$g_r(t):=(1-r\cos t)(1-r\sin t)\qquad(0\leq t\leq{\pi\over2})$$ we find the following (see the figure):

(a) When $r<\rho:=2(\sqrt{2}-1)$ then $$q_1q_2>0\quad\Rightarrow\quad f(q_1,q_2)<f(r,0)\ .$$ (b) When $r>\rho$ then $$q_1\ne q_2\quad\Rightarrow\quad f(q_1,q_2)<f\left({r\over\sqrt{2}},{r\over\sqrt{2}}\right)\ .$$ ${\bf 2\ }$ Now let $n\geq2$ be arbitrary and assume that $q=(q_1,q_2,\ldots,q_n)$ is an admissible point where $f$ assumes its maximum. Put $$\tau:={1\over2}\rho^2\doteq0.343>{1\over3}\ .$$ In the light of ${\bf 1}$ there can be at most one $i$ with $0<q_i^2<\tau$, and as $\tau>{1\over3}$ there can be at most two $i$ with $q_i^2\geq\tau$. All other $q_i$ have to be zero.

${\bf 3\ }$ When $0<q_1^2<\tau\leq q_2^2\leq q_3^2$ then $$q_1^2+q_2^2=1-q_3^2\leq 1-\tau<2\tau=\rho^2\ .$$ From ${\bf 1}$(a) we then conclude that $q$ cannot be a maximum point.

${\bf 4\ }$ When only $q_1^2$ and $q_2^2$ are $>0$ then ${\bf 1}$(b) with $r=1$ tells us that we have $q_1=q_2={1\over\sqrt{2}}$, and this leads to the maximal value $$\max f=\left(1-{1\over\sqrt{2}}\right)^2={1\over2}(3-2\sqrt{2})\doteq0.0858\ .$$

Addendum: I've included the full set of solutions Mathematica finds for the case $n=3$ at the end of this answer.

The comments and answers so far don't show how Lagrange multipliers can solve the problem (because of the non-negativity condition). Here's how that technique can be used.

To maximize a function $f(q_1,\dots, q_n)$ with the requirement that $q_i\ge0$ and an additional constraint $c(q_1,\dots, q_n)=0$, it suffices to maximize the function $f(s_1^2,\dots, s_n^2)$ subject to the constraint $c(s_1^2,\dots, s_n^2)=0$, then choose those solutions with all $s_i$ real that give the largest value of $f$. For these, $(s_1^2,\dots, s_n^2)$ is a non-negative solution to the original constrained maximization question.

For this particular question, then, maximize $\prod(1-s_i^2)$ subject to $\sum s_i^4=1$.

Here’s the three-variable case, where we need to maximize $(1-a)(1-b)(1-c)$ for $a,b,c\ge0$, subject to $a^2+b^2+c^2=1$.

Maximize $(1-x^2)(1-y^2)(1-z^2)$ subject to $x^4+y^4+z^4=1$. Using Lagrange multipliers, let $g(x,y,z,\lambda)=(1-x^2)(1-y^2)(1-z^2)-\lambda(x^4+y^4+z^4-1)$, and solve the following system of equations over the real numbers:

$$\begin{align} 0 = \frac{\partial f}{\partial x} = & -2x(1-y^2)(1-z^2)+4x^3\lambda\\ 0 = \frac{\partial f}{\partial y} = & -2y(1-x^2)(1-z^2)+4y^3\lambda\\ 0 = \frac{\partial f}{\partial z} = & -2z(1-x^2)(1-y^2)+4z^3\lambda\\ 0 = \frac{\partial f}{\partial \lambda} = & x^4+y^4+z^4-1\\ \end{align}$$

When a solution $(x,y,z,\lambda)$ of this system gives a maximum among all real solutions, $(x^2,y^2, z^2)$ is a solution to the original question.

The system is no fun to solve, and I didn't try to generalize to see if it gives a nice answer the original question in general. In theory it will, but whether it's any simpler than the other answers, I don't know.

Here are all of Mathematica’s real solutions to the system above, written in terms of $a=x^2,b=y^2,c=z^2$.

$ \begin{array}{llll} (a,b,c) & (1-a)(1-b)(1-c)\\ \hline\\ \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) & \left(1-\frac{1}{\sqrt{2}}\right)^2 \\ \left(\frac{1}{\sqrt{2}} , 0 , \frac{1}{\sqrt{2}}\right) & \left(1-\frac{1}{\sqrt{2}}\right)^2 \\ \left(0 , \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right)& \left(1-\frac{1}{\sqrt{2}}\right)^2 \\ \left(\frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} \right)& \left(1-\frac{1}{\sqrt{3}}\right) ^3 \\ \left(\frac{2}{3}, \frac{2}{3} , \frac{1}{3} \right)& \frac{2}{27} \\ \left(\frac{2}{3} , \frac{1}{3} , \frac{2}{3} \right)& \frac{2}{27} \\ \left(\frac{1}{3} , \frac{2}{3} , \frac{2}{3} \right)& \frac{2}{27} \\ (1 , 0 , 0) & 0 \\ (0 , 1 , 0) & 0 \\ (0 , 0 , 1) & 0 \\ \end{array} $