Optimization without using second derivatives I have the following maximization problem:
$$\max_{x_1, \dots, x_{n+1}} x_{n+1}\left(a-b\sum_{k=1}^n \frac{1}{x_k}\right)$$
subject to the equality constraint
$$x_{n+1}=1-2M\sum_{k=1}^n x_{k}$$ 
and the non-negativity constraints $x_k \ge 0$ for $k=1,\dots,n+1$. In this problem, $a$ and $b$ are positive constants and $M$ is an integer. 
I want to prove that the point 
$$x_k=x^*=\sqrt{\frac{b}{2Ma}}$$ 
for all $k=1,\dots,n$ and $x_{n+1}=1-2Mnx^*$ is the maximum. Here I suppose that constant are such as $1-2Mnx^*>0$.
To get this point I supposed $x_k>0$ for $k=1,\dots,n+1$ and I used KKT theorem which give to me an unique KKT-point. As the value at this point is positive, this point must to be a maximum (I think).
I want to prove this formally without use second derivates. However, I dont know how to proced because function is not bounded near to $x_k=0$. 
Thanks for help!. 
 A: Having various constant in such a problem makes it a bit more difficult to see through it. Although, strictly speaking it is not really necessary, we can simplify the appearance of the problem by using a simple transformation. In particular we can use $x_i = \frac{y_i}{f}$ for $i \leq n$ and some yet unknown constant $f$. Then we can rewrite the problem as:
$$
\max_{\{y_i\}} \left( 1 - \frac{2 M}{f} \sum_{i=1}^n y_i\right)\left( a - b f \sum_{i=1}^n \frac{1}{y_i}\right)
$$
with $y_i>0$ and both factors are required to be positive. We can now extract some of the constant and we obtain
$$
\max_{\{y_i\}} ~(2 M b) \left( \frac{f}{2 M} - \sum_{i=1}^n y_i\right)\left( \frac{a}{b f} - \sum_{i=1}^n \frac{1}{y_i}\right)
$$
By conveniently choosing $f=\sqrt{\frac{2 M a}{b}}$ and defining $q=\sqrt{\frac{a}{2 M b}}$ we see that the optimisation problem actually only depends on a single constant.
$$
\max_{\{y_i\}} ~(2 M b) \left(q - \sum_{i=1}^n y_i\right)\left( q - \sum_{i=1}^n \frac{1}{y_i}\right)
$$
If we apply the fact that the arithmetic mean is larger equal to the harmonic mean $\frac{1}{n} \sum_i a_i \geq n \left( \sum_i \frac{1}{a
_i}\right)^{-1}$, in particular for $a_i=\frac{1}{y_i}$ we get
$$
\max_{\{y_i\}} ~(2 M b) \left(q - \sum_{i=1}^n y_i\right)\left( q - \sum_{i=1}^n \frac{1}{y_i}\right) \leq \max_{\{y_i\}} ~(2 M b) \left(q - \sum_{i=1}^n y_i\right)\left( q - \frac{n^2}{\sum_{i=1}^n y_i}\right) \\
= 2 M b \left( q - n \bar{y}\right) \left(q - \frac{n}{\bar{y}} \right) \\
= 2 M b \left( \left(n-q\right)^2 - \frac{n q}{\bar{y}} \left(\bar{y}-1 \right)^2\right)\\
\leq 2 M b \left(n-q\right)^2
$$
where $\bar{y} = \frac{1}{n} \sum_{i=1}^n y_i$ and equality only occurs when $y_i=y_j$ for all $i,j$ for the inequality of the means and $\bar{y}=1$ in the last step. Hence the maximum is obtained for $x_i = \frac{1}{f} = \sqrt{\frac{b} {2M a}}$ for all $i \leq n$, just as you expected.
There is however the constraint on the constants to guarantee that both factors in the maximum are positive and that this solutions is permissible, which is the case whenever $a > 2 b n^2 M$.  
