Showing that for real $0<x_i, 0<q$ with $x_1…x_n=q^n$ it holds that $(1+x_1)…(1+x_n)\geq(1+q)^n$

Using method of Lagrange multipliers, I am looking to minimise the function

$f(x_1,...,x_n)=\prod_{i=1}^n(1+x_i)$ with the side condition $g(x_1,...,x_n)=\prod_{i=1}^nx_i-q^n=0$.

My goal is to show that $f$ is minimal when $x_i=q$ for all $i$.

I have $$\nabla g=(\prod_{i=1, i \neq j}^nx_i)_{1\leq j \leq n}$$ $$\nabla f=(\prod_{i=1, i \neq j}^n(1+x_i))_{1\leq j \leq n}$$

Now I know that $\nabla f = \lambda \nabla g$, so for all $j$:

$$\prod_{i=1, i \neq j}^n(1+x_i)=\lambda\prod_{i=1, i \neq j}^nx_i$$

and since all $x_i>0$

$$\prod_{i=1, i \neq j}^n(1+\frac{1}{x_i})=\lambda$$

Since the products are equal for all $j$, it follows that the $x_1=x_2=...=x_n$ and thus $x_1...x_n=x_1^n=q^n \iff x_i=q$ for all $i$. So far so good.

My problem is to show that it is a minimum: The entries of the Hessian are

$$\partial_{x_i} ^2f=0$$

$$\partial_{x_j} \partial_{x_k}f=\prod_{i=1, i \neq j,i \neq k}^n(1+x_i)>0$$

So I have a matrix with zero diagonal and positive entries everywhere else. This matrix is not positive definite, as is easily seen in the $2 \times 2$ case.

Did I do any mistakes? How do I argue that it is a minimum?

• You could also try HUYGEN’S INEQUALITY, stating that for $x_i\geq0$ $$(1+x_1)(1+x_2)...(1+x_n)\geq\left(1+\left(x_1x_2...x_n\right)^{\frac{1}{n}}\right)^n$$ – rtybase Jul 23 '18 at 21:49
• Even if the Hessian would signalize a local minimum at $x_i=q$ $(i\in[n])$ you could not be sure that it is actually the global minimum. – Christian Blatter Jul 24 '18 at 9:09

The associated Bordered Hessian matrix is $$\left[ \begin{array}{c|ccc} 0 & (1+q)^{n-1} & (1+q)^{n-1} & (1+q)^{n-1} &\dots \\\hline (1+q)^{n-1} & 0 & (1+q)^{n-2} & (1+q)^{n-2} &\dots \\ (1+q)^{n-1} & (1+q)^{n-2} & 0 & (1+q)^{n-2} &\dots \\ (1+q)^{n-1} & (1+q)^{n-2} & (1+q)^{n-2} & 0 &\dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right]$$ Up to a multiplicative constant, the above has the shape $$A = \left[ \begin{array}{c|cccc} 0 & a & a & a &\dots & a \\\hline a & 0 & 1 & 1 &\dots & 1\\ a & 1 & 0 & 1 &\dots & 1\\ a & 1 & 1 & 0 &\dots & 1\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ a &1 & 1 & 1 & \dots & 0 \end{array} \right]\ .$$ It is an $(n+1)\times(n+1)$ matrix, and its characteristic polynomial is $$P_A(x) = (x+1)^{n-1}(x^2-(n-1)x-na^2)\ .$$ So exactly one eigenvalue is positive. Let $C$ be the matrix: $$C= \left[\begin{array}{c|cccc} 1 & 0 & 0 & 0 & \dots & 0 \\\hline 0 & 1 & 0 & 0 & \ddots & 0 \\ 0 & -1 & 1 & 0 & \ddots & 0 \\ 0 & 0 & -1 & 1 & \ddots & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & -1 & 1 \end{array}\right]\ .$$ ($C$ has ones on the diagonal, in the $n\times n$ block "minus ones" immediately under the diagonal, else zeros.) Then conjugation with $C$ gives: $$CAC^{-1} = \left[\begin{array}{c|c|ccc} 0 & na & (n-1)a & (n-2)a & \dots & a \\\hline a & n-1 & (n-1) & (n-2) & \dots & 1 \\\hline 0 & 0 & -1 & 0 & \dots & 0 \\ 0 & 0 & 0 & -1 & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \dots & -1 \end{array}\right]\ .$$ As as a quadratic form, $A$ changes by the base change action of $C$ into $$CAC^t = \left[ \begin{array}{cc|cccc} 0 & a & 0 & 0 & 0 &\dots & 0\\ a & 0 & 1 & 0 & 0 &\ddots & 0\\\hline 0 & 1 & -2 & 1 & 0 &\ddots & 0\\ 0 & 0 & 1 & -2 & 1 &\ddots & 0\\ 0 & 0 & 0 & 1 & -2 &\ddots & 0\\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\ 0 & 0 & 0 & 0 & 0 & \dots & -2 \end{array} \right]\ .$$

If Lagrange multiplicators are used, the optimization problem is not regarging the "full function $f+\lambda g$", and the Hessian matrix of $f+\lambda g$ may fulfill sufficient positivity / negativity conditions for only a "tangential null space". The idea is roughly to have a "conditional Taylor polynomial of order two in the $x$-variables", that still allows to deduce a local extremal value.

In our case, the condition is $x_1x_2\dots x_n=q^n$. Formally, writing $x_1=q+h_1+O(h_1^2)$ and analogously for the other variables, we get the relation $$\prod_{1\le k\le n}(q+h_k+O(h_k^2))=q^n\ ,$$ so formally $q^{n-1}(h_1+h_2+\dots+h_n)+O(|h|^2)=0$.

In the given sitation, the matrix $C$ provides an $(n-1)\times (n-1)$ block which is built with vectors from the null space. Eventually, one can work to convert this beginning into a proof.

The most simple solution to the minimum problem is to observe that if two component values in $x=(x_1,x_2,x_3,\dots)$ are not equal, say $x_1\ne x_2$ without loss of generality, then we can redistribute $x_1x_2=c^2$ in a new point $x_c:=(c,c,x_3,\dots)$ and we get a smaller value because $$(1+x_1)(1+x_2)-(1+c)^2=x_1+x_2-2c=(\sqrt x_1-\sqrt x_2)^2>0\ .$$ (Because of this simpler argument i did not insist to complete the proof using Lagrange multiplicators. Search please the net for the bordered matrix to see many explicit examples.)

Considering the Lagrangian

$$L(x,\lambda) = \prod_k^n(x_k+1)-(q+1)^n+\lambda\left(\prod_k^n x_k - q^n\right)$$

we have the stationary conditions

$$L_{x_k} = \prod_{j\ne k}^n(x_j+1)+\lambda\prod_{j\ne k}^n x_j = 0$$

or

$$\lambda = -\frac{\prod_{j\ne k}^n(x_j+1)}{\prod_{j\ne k}^n x_j}$$

hence

$$\lambda = -\frac{\prod_{j\ne k}^n(x_j+1)}{\prod_{j\ne k}^n x_j} = -\frac{\prod_{i\ne k}^n(x_i+1)}{\prod_{i\ne k}^n x_i}\Rightarrow \frac{x_j}{x_j+1} = \frac{x_i}{x_i+1}\Rightarrow x_1=x_2=\cdots=x_n = q$$

This stationary point is a saddle point for $\prod_k^n(x_k+1)-(q+1)^n$ as can be checked easily analyzing the behavior of

$$(q+\epsilon+1)^n-(q+1)^n$$

for $\epsilon \in [-1,1]$

This is not a problem because the qualification should be done with the Hessian for

$$F(x)=\left(\prod_k^n(x_k+1)-(q+1)^n\right)\circ \left(\prod_k^n x_k - q^n\right)$$

I leave it here in the hope of finding a suitable expression for such hessian.

NOTE

For the case $n = 3$ we have

$$\left((x+1)(y+1)(z+1)-(q+1)^3\right)\circ\left(z=\frac{q^3}{x y}\right) = (x+1) (y+1) \left(\frac{q^3}{x y}+1\right)-(q+1)^3$$

with Hessian $H$ evaluated at $x=y=z=q$

$$H = \left( \begin{array}{cc} 2+\frac{2}{q} & 1+\frac{1}{q} \\ 1+\frac{1}{q} & 2+\frac{2}{q} \\ \end{array} \right)$$

with eigenvalues

$$\left\{3 \left(\frac{1}{q}+1\right),\frac{1}{q}+1\right\}$$

characterizing a minimum.

• Why is the Hessian of the composite function the important one for this problem? – B.Swan Jul 24 '18 at 14:52
• I did not know that one has to be consider the Bordered Hessian! – B.Swan Jul 24 '18 at 15:01

Proof

Apply Carison's inequality, which also could be viewed as a generalized form of Cauchy-Schwarz inequality

$$(x_1+y_1+\cdots)(x_2+y_2+\cdots)\cdots(x_n+y_n+\cdots)\geq \left[\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}+\left(\prod_{i=1}^n y_i\right)^{\frac{1}{n}}+\cdots\right]^n,$$where $x_i,y_i,\cdots\geq0$ for $i=1,2,\cdots$

Thus, $$(1+x_1)\cdots(1+x_n)\geq \left[\left(\prod_{i=1}^n 1\right)^{\frac{1}{n}}+\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\right]^{n}=\left[1+(q^n)^{\frac{1}{n}}\right]^n=(1+q)^n.$$

Since an answer with the Lagranian method has been given, here is an alternative proof using the Weighted AM-GM Inequality. Note that $$\frac{q}{1+q}\,\left(1+x_i\right)=\frac{1}{1+q}\cdot q+\frac{q}{1+q}\cdot x_i\geq q^{\frac{1}{1+q}}\,x_i^{\frac{q}{1+q}}\text{ for }i=1,2,\ldots,n\,.$$ Therefore, $$1+x_i\geq (1+q)\,\left(\frac{x_i}{q}\right)^{\frac{q}{1+q}}\text{ for }i=1,2,\ldots,n\,.$$ Consequently, $$\prod_{i=1}^n\,\left(1+x_i\right)\geq (1+q)^n\,\left(\frac{\prod\limits_{i=1}^n\,x_i}{q^n}\right)^{\frac{q}{1+q}}=(1+q)^n\,.$$