# Showing the maximum value using alternate approach

If $$x_1,x_2...x_n$$ are postive numbers satisfying $$x_1\cdot x_2 \cdots x_n = 1$$ , then find the maximum value of $$\frac{1}{\sqrt{x_1 ^2 + 1}\,\cdot\, \sqrt{x_2^2 +1}\,\cdots\,\sqrt{x_n^2+1} }$$ .

What i considered was applying the RMS -GM inequality which is $$\sqrt{a^2 +b^2} \geq 2\sqrt{ab}$$ which gives the maximum of the expression but is there an another way which can be used to systematically show and get the maximum ?

• Note : originally problem was given in terms of $$\cot z_1 \cdots \cot z_n = 1$$ and all angles lie in $$(0,π/2)$$ and we need to find maximum of $$\cos z_1\cdots\cos z_n$$ but for better understanding i converted into this form
• The problem has a full symmetry over variables $x_1,...., x_n$. It means that we can check the case $x_1=x_2=... =x_n$ first. We can see that it is the maximum. Indeed, let's take $x_1=x_2=...=x_{n-2}=1$ and $x_{n-1}=\frac{1}{x_n}$; then, leading $x_n\to 0$, we can get as small value as we want. Jun 24, 2022 at 17:52
• I understood most but how just checxking the x_1 = x_2 = ... Case one concludes its the maximum ? @Svyatoslav Jun 24, 2022 at 18:00

We can define $$\displaystyle f(x_1, ..., x_n)=\ln \frac{1}{\sqrt{x_1 ^2 + 1}\,\cdot\, \sqrt{x_2^2 +1}\,\cdots\,\sqrt{x_n^2+1} }=-\frac{1}{2}\Big(\ln(x_1^2+1)+...+\ln(x_n^2+1)\Big)$$.
As logarithm is a growing function, it is enough to find the maximum of $$f(x)$$. Given that $$\ln(x_1...x_n)=0$$, we also introduce the additional condition via $$g(x_1, ..., x_n)=\ln(x_1...x_n)=\ln x_1+... \ln x_n$$, and consider the Lagrangian function $$L(x_1, ... x_n)=f(x_1, ..., x_n)+\lambda g(x_1, ..., x_n)$$ We we are looking for an unconditional extremum of $$L(x_1, ... x_n)$$: $$\frac{\partial}{\partial x_i}L(x_1, ... x_n)=0\,\,\Rightarrow\,\,\frac{\lambda}{x_i}-\frac{x_i}{x_i^2+1}=0\,\,\Rightarrow\,\,x_i^2=\frac{1-\lambda}{\lambda}$$ From the condition $$\displaystyle g(x_1, ..., x_n)=\ln x_1+... \ln x_n=0$$ we get $$2n\ln\frac{1-\lambda}{\lambda}=0\,\,\Rightarrow\,\,\lambda=\frac{1}{2}\,\,\Rightarrow\,\, \,x_1=x_2=...=x_n=1$$ We have a single conditional extremum; taking, for example, $$x_1=...x_{n-2}=1,\,x_{n-1}=\frac{1}{x_n}$$ and leading $$x_n\to 0$$ (and $$x_{n-1}\to\infty$$ correspondingly), we can make $$f(x)$$ as small as we wish; therefore, this extremum is a maximum.