A stationary point of a function $f$ is a real number $r$ such that $f ^ { \prime } ( r ) = 0$. A polynomial need not have a stationary point (e.g. $x ^ { 3 } + x$ has none). Consider a polynomial $p ( x )$.
(a) If $p ( x )$ is of degree 2022, then $p ( x )$ must have at least one stationary point.
(b) If the number of distinct real roots of $p ( x )$ is 2021, then $p ( x )$ must have at least 2020 stationary points.
(c) If the number of distinct real roots of $p ( x )$ is 2021, then $p ( x )$ can have at most 2020 stationary points.
(d) If $r$ is a stationary point of $p ( x )$ AND $p ^ { \prime \prime } ( r ) = 0$, then the point $( r , p ( r ) )$ is neither a local maximum nor a local minimum point on the graph of $p ( x )$.

A function $f ( x )$ defined for $x > a$ and a quartic function $g ( x )$ with leading coefficient $-1$ satisfy the following conditions. (Here, $a$ is a constant.)
(a) For all real numbers $x > a$, $$( x - a ) f ( x ) = g ( x ).$$ (b) For two distinct real numbers $\alpha , \beta$, the function $f ( x )$ has the same local maximum value $M$ at $x = \alpha$ and $x = \beta$. (Here, $M > 0$)
(c) The number of values of $x$ where the function $f ( x )$ has a local maximum or minimum is greater than the number of values of $x$ where the function $g ( x )$ has a local maximum or minimum. When $\beta - \alpha = 6 \sqrt { 3 }$, find the minimum value of $M$. [4 points]

We assume that $f$ is the zero application on $C(0,1)$ and that $u$ is an element of $\mathcal{D}_f$. For all $n \in \mathbb{N}$, we define the application $$u_n : \begin{array}{rll} \bar{D}(0,1) & \rightarrow & \mathbb{R} \\ (x,y) & \mapsto & u(x,y) + \dfrac{1}{n}(x^2 + y^2) \end{array}$$
Suppose that $u_n$ admits a local maximum at $(\tilde{x}, \tilde{y}) \in D(0,1)$.
a) By examining the behavior of the function $x \mapsto u_n(x, \tilde{y})$ show that, in this case, $\partial_{11} u_n(\tilde{x}, \tilde{y}) \leqslant 0$. Similarly, one can show that $\partial_{22} u_n(\tilde{x}, \tilde{y}) \leqslant 0$. Thus $\Delta u_n(\tilde{x}, \tilde{y}) \leqslant 0$. This result is admitted for the rest.
b) Deduce that $u_n$ does not admit a local maximum on $D(0,1)$.

Let $U$ be a non-empty bounded open set of $\mathbb{R}^n$ ($n \geqslant 2$) and $f : U \rightarrow \mathbb{R}$ of class $\mathcal{C}^2$. Let $f$ be a function continuous on $\bar{U}$. Show that $f$ attains a maximum at some point $x_0 \in \bar{U}$.

Let $a < b$ be two real numbers and $f : [a,b] \rightarrow \mathbb{R}$ be an infinitely differentiable function. Let us call (H) the following hypothesis: there exists a unique point $x_0 \in [a,b]$ where $f$ attains its maximum, we have $a < x_0 < b$, and $f''(x_0) \neq 0$.
Show that under hypothesis (H), we have $f''(x_0) < 0$.

From now on, $f$ denotes an infinitely differentiable function from $[0,1]$ to $\mathbb{R}$. We assume that there exists a unique point $x_0 \in [0,1]$ where $f'$ vanishes. We also assume that $f''(x_0) > 0$.
For all $x \in [x_0, 1]$, we define $$h(x) = \sqrt{|f(x) - f(x_0)|}$$
(a) Show that the function $h$ defines a bijection from $[x_0, 1]$ to $[0, h(1)]$.
(b) Show that the application $h$ is differentiable at $x_0$ from the right, and that $h'_+(x_0) = \sqrt{\frac{f''(x_0)}{2}}$.

Let $a < b$ be two real numbers and $f : [ a , b ] \rightarrow \mathbb { R }$ be an infinitely differentiable function. Let us call (H) the following hypothesis: there exists a unique point $x _ { 0 } \in [ a , b ]$ where $f$ attains its maximum, we have $a < x _ { 0 } < b$, and $f ^ { \prime \prime } \left( x _ { 0 } \right) \neq 0$.
Show that under hypothesis $( H )$, we have $f ^ { \prime \prime } \left( x _ { 0 } \right) < 0$.

From now on, $f$ denotes an infinitely differentiable function from $[ 0,1 ]$ to $\mathbb { R }$. We assume that there exists a unique point $x _ { 0 } \in \left[ 0,1 \left[ \right. \right.$ where $f ^ { \prime }$ vanishes. We also assume that $f ^ { \prime \prime } \left( x _ { 0 } \right) > 0$. We are also given an infinitely differentiable function $g : [ 0,1 ] \rightarrow \mathbb { R }$.
For all $x \in \left[ x _ { 0 } , 1 \right]$, we define $$h ( x ) = \sqrt { \left| f ( x ) - f \left( x _ { 0 } \right) \right| }$$
(a) Show that the function $h$ defines a bijection from $\left[ x _ { 0 } , 1 \right]$ to $[ 0 , h ( 1 ) ]$.
(b) Show that the map $h$ is differentiable at $x _ { 0 }$ on the right, and that $h ^ { \prime } \left( x _ { 0 } \right) = \sqrt { \frac { f ^ { \prime \prime } \left( x _ { 0 } \right) } { 2 } }$.

Let $p \in \mathbb{N}^{\star}$. We assume that $f, g_1, \ldots, g_p$ are functions from $\mathbb{R}^n$ to $\mathbb{R}$ differentiable on $\mathbb{R}^n$, and that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. For all $x \in K$, we denote $I_x = \left\{ i \in \llbracket 1, p \rrbracket, g_i(x) = 0 \right\}$. Show that for all $x \in K$, $$\mathcal{A}_K(x) \subset \left\{ h \in \mathbb{R}^n, \forall i \in I_x, \langle \nabla g_i(x), h \rangle \leqslant 0 \right\}.$$

Let $p \in \mathbb{N}^{\star}$. We assume that $f, g_1, \ldots, g_p$ are functions from $\mathbb{R}^n$ to $\mathbb{R}$ differentiable on $\mathbb{R}^n$, and that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. For all $x \in K$, we denote $I_x = \left\{ i \in \llbracket 1, p \rrbracket, g_i(x) = 0 \right\}$. We consider $x^{\star} \in K$ and we make the following hypothesis: $$(H) \quad \text{there exists } v \in \mathbb{R}^n \text{ such that for all } i \in I_{x^{\star}}, \langle \nabla g_i(x^{\star}), v \rangle < 0.$$ Show that $\mathcal{A}_K(x^{\star}) = \left\{ h \in \mathbb{R}^n, \forall i \in I_{x^{\star}}, \langle \nabla g_i(x^{\star}), h \rangle \leqslant 0 \right\}$.

Let $p \in \mathbb{N}^{\star}$. We assume that $f, g_1, \ldots, g_p$ are functions from $\mathbb{R}^n$ to $\mathbb{R}$ differentiable on $\mathbb{R}^n$, and that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. For all $x \in K$, we denote $I_x = \left\{ i \in \llbracket 1, p \rrbracket, g_i(x) = 0 \right\}$. Show that if $x^{\star} \in K$ is such that $(\nabla g_i(x^{\star}))_{i \in I_{x^{\star}}}$ forms a free family, then hypothesis $(H)$ is verified.

Let $p \in \mathbb{N}^{\star}$. We assume that $f, g_1, \ldots, g_p$ are functions from $\mathbb{R}^n$ to $\mathbb{R}$ differentiable on $\mathbb{R}^n$, and that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. For all $x \in K$, we denote $I_x = \left\{ i \in \llbracket 1, p \rrbracket, g_i(x) = 0 \right\}$. Suppose that $f$ attains at $x^{\star} \in K$ a local minimum on $K$, and that hypothesis $(H)$ is verified. Show that there exist non-negative real numbers $\mu_1^{\star}, \ldots, \mu_p^{\star}$ such that $$\left\{ \begin{array}{l} \nabla f(x^{\star}) + \sum_{i=1}^{p} \mu_i^{\star} \nabla g_i(x^{\star}) = 0 \\ \mu_i^{\star} g_i(x^{\star}) = 0 \text{ for all } i \in \llbracket 1, p \rrbracket. \end{array} \right.$$

Let $p \in \mathbb{N}^{\star}$. We assume that $f, g_1, \ldots, g_p$ are functions from $\mathbb{R}^n$ to $\mathbb{R}$ differentiable on $\mathbb{R}^n$, and that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. Suppose in this question that the functions $f, g_1, \ldots, g_p$ are convex. Let $x^{\star} \in K$ and $\mu_1^{\star}, \ldots, \mu_p^{\star} \in \mathbb{R}_+$ be such that $$\left\{ \begin{array}{l} \nabla f(x^{\star}) + \sum_{i=1}^{p} \mu_i^{\star} \nabla g_i(x^{\star}) = 0 \\ \mu_i^{\star} g_i(x^{\star}) = 0 \text{ for all } i \in \llbracket 1, p \rrbracket \end{array} \right.$$ is verified. Show that $f$ admits at $x^{\star}$ a global minimum on $K$.

Let $p \in \llbracket 1, n \rrbracket$. We assume that $f, g_1, \ldots, g_p$ are differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}$, that $f$ is $\alpha$-convex for some $\alpha \in \mathbb{R}_+^{\star}$, and that the functions $g_1, \ldots, g_p$ are convex. We further assume that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. We denote $g(x) = \begin{pmatrix} g_1(x) \\ \vdots \\ g_p(x) \end{pmatrix}$ for all $x \in \mathbb{R}^n$. We introduce the function $\mathcal{L} : \mathbb{R}^n \times \mathbb{R}_+^p \rightarrow \mathbb{R}$ defined by $$\mathcal{L}(x, \mu) = f(x) + \sum_{i=1}^{p} \mu_i g_i(x)$$ for all $x \in \mathbb{R}^n$ and all $\mu = (\mu_1, \ldots, \mu_p) \in \mathbb{R}_+^p$. Show that $\inf_{x \in K} f(x) = \inf_{x \in \mathbb{R}^n} \sup_{\mu \in \mathbb{R}_+^p} \mathcal{L}(x, \mu)$.

Let $p \in \llbracket 1, n \rrbracket$. We assume that $f, g_1, \ldots, g_p$ are differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}$, that $f$ is $\alpha$-convex for some $\alpha \in \mathbb{R}_+^{\star}$, and that the functions $g_1, \ldots, g_p$ are convex. We further assume that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. We denote $g(x) = \begin{pmatrix} g_1(x) \\ \vdots \\ g_p(x) \end{pmatrix}$ for all $x \in \mathbb{R}^n$. We introduce the function $\mathcal{L} : \mathbb{R}^n \times \mathbb{R}_+^p \rightarrow \mathbb{R}$ defined by $$\mathcal{L}(x, \mu) = f(x) + \sum_{i=1}^{p} \mu_i g_i(x)$$ for all $x \in \mathbb{R}^n$ and all $\mu = (\mu_1, \ldots, \mu_p) \in \mathbb{R}_+^p$. Show that for all $\mu \in \mathbb{R}_+^p$, there exists a unique $x_\mu \in \mathbb{R}^n$ satisfying $\mathcal{L}(x_\mu, \mu) = \inf_{x \in \mathbb{R}^n} \mathcal{L}(x, \mu)$.

Let $p \in \llbracket 1, n \rrbracket$. We assume that $f, g_1, \ldots, g_p$ are differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}$, that $f$ is $\alpha$-convex for some $\alpha \in \mathbb{R}_+^{\star}$, and that the functions $g_1, \ldots, g_p$ are convex. We further assume that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. We denote $g(x) = \begin{pmatrix} g_1(x) \\ \vdots \\ g_p(x) \end{pmatrix}$ for all $x \in \mathbb{R}^n$. We introduce the function $\mathcal{L} : \mathbb{R}^n \times \mathbb{R}_+^p \rightarrow \mathbb{R}$ defined by $$\mathcal{L}(x, \mu) = f(x) + \sum_{i=1}^{p} \mu_i g_i(x)$$ for all $x \in \mathbb{R}^n$ and all $\mu = (\mu_1, \ldots, \mu_p) \in \mathbb{R}_+^p$. For all $\mu \in \mathbb{R}_+^p$, we denote $G(\mu) := \inf_{x \in \mathbb{R}^n} \mathcal{L}(x, \mu) = \mathcal{L}(x_\mu, \mu)$. We say that $(\bar{x}, \bar{\mu}) \in \mathbb{R}^n \times \mathbb{R}_+^p$ is a saddle point of $\mathcal{L}$ if $$\mathcal{L}(\bar{x}, \bar{\mu}) = \inf_{x \in \mathbb{R}^n} \mathcal{L}(x, \bar{\mu}) \quad \text{and} \quad \mathcal{L}(\bar{x}, \bar{\mu}) = \sup_{\mu \in \mathbb{R}_+^p} \mathcal{L}(\bar{x}, \mu).$$ We assume in this question that $(\bar{x}, \bar{\mu}) \in \mathbb{R}^n \times \mathbb{R}_+^p$ is a saddle point of $\mathcal{L}$. a. Show that $\bar{x}$ is a solution of $(P)$: $\inf_{x \in K} f(x)$. b. Show that $\bar{\mu}$ is a solution of $(Q)$: $\sup_{\mu \in \mathbb{R}_+^p} G(\mu)$. c. Show that $\inf_{x \in \mathbb{R}^n} \sup_{\mu \in \mathbb{R}_+^p} \mathcal{L}(x, \mu) = \sup_{\mu \in \mathbb{R}_+^p} \inf_{x \in \mathbb{R}^n} \mathcal{L}(x, \mu)$.

Let $p \in \llbracket 1, n \rrbracket$. We assume that $f, g_1, \ldots, g_p$ are differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}$, that $f$ is $\alpha$-convex for some $\alpha \in \mathbb{R}_+^{\star}$, and that the functions $g_1, \ldots, g_p$ are convex. We further assume that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. We denote $g(x) = \begin{pmatrix} g_1(x) \\ \vdots \\ g_p(x) \end{pmatrix}$ for all $x \in \mathbb{R}^n$. We introduce the function $\mathcal{L} : \mathbb{R}^n \times \mathbb{R}_+^p \rightarrow \mathbb{R}$ defined by $$\mathcal{L}(x, \mu) = f(x) + \sum_{i=1}^{p} \mu_i g_i(x)$$ for all $x \in \mathbb{R}^n$ and all $\mu = (\mu_1, \ldots, \mu_p) \in \mathbb{R}_+^p$. For all $\mu \in \mathbb{R}_+^p$, we denote $G(\mu) := \inf_{x \in \mathbb{R}^n} \mathcal{L}(x, \mu)$. We consider $x^{\star} \in K$ a solution of $(P)$ satisfying hypothesis $(H)$. Let $\mu^{\star} = (\mu_1^{\star}, \ldots, \mu_p^{\star})$ as in question III.7. Show that $\mu^{\star}$ is a solution of $(Q)$.

Let $p \in \llbracket 1, n \rrbracket$. We assume that $f, g_1, \ldots, g_p$ are differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}$, that $f$ is $\alpha$-convex for some $\alpha \in \mathbb{R}_+^{\star}$, and that the functions $g_1, \ldots, g_p$ are convex. We further assume that $$K = \left\{ x \in \mathbb{R}^n, g_1(x) \leqslant 0, \ldots, g_p(x) \leqslant 0 \right\}$$ is non-empty. We denote $g(x) = \begin{pmatrix} g_1(x) \\ \vdots \\ g_p(x) \end{pmatrix}$ for all $x \in \mathbb{R}^n$. We introduce the function $\mathcal{L} : \mathbb{R}^n \times \mathbb{R}_+^p \rightarrow \mathbb{R}$ defined by $$\mathcal{L}(x, \mu) = f(x) + \sum_{i=1}^{p} \mu_i g_i(x)$$ for all $x \in \mathbb{R}^n$ and all $\mu = (\mu_1, \ldots, \mu_p) \in \mathbb{R}_+^p$. For all $\mu \in \mathbb{R}_+^p$, we denote $G(\mu) := \inf_{x \in \mathbb{R}^n} \mathcal{L}(x, \mu) = \mathcal{L}(x_\mu, \mu)$. We assume throughout this question that the function $\mu \in \mathbb{R}_+^p \mapsto x_\mu$ is continuous. We consider a solution $\bar{\mu} \in \mathbb{R}_+^p$ of $(Q)$. a. Let $\mu \in \mathbb{R}_+^p$ and $\xi \in \mathbb{R}^p$ be such that $\mu + \xi \in \mathbb{R}_+^p$. Show that for all $t \in [0,1]$, $\mu + t\xi \in \mathbb{R}_+^p$, and $$\lim_{\substack{t \rightarrow 0 \\ t > 0}} \frac{G(\mu + t\xi) - G(\mu)}{t} = \langle g(x_\mu), \xi \rangle.$$ Deduce that for all $\mu \in \mathbb{R}_+^p$, $\langle g(x_{\bar{\mu}}), \mu - \bar{\mu} \rangle \leqslant 0$. b. Show that $x_{\bar{\mu}}$ is a solution of $(P)$.

We denote $\|\cdot\|_2$ the canonical Euclidean norm on $\mathbb{R}^2$ and we denote $$\mathcal{C} := \left\{ x \in \mathbb{R}^2 \mid \|x\|_2 = 1 \right\}$$ We fix an arbitrary norm $\|\cdot\|$ on $\mathbb{R}^2$ and we denote $$\mathcal{K} = \left\{ A \in M_2(\mathbb{R}) \mid \forall x \in \mathbb{R}^2,\ \|x\|_2 \geq \|Ax\| \right\}.$$ a) Show that $\mathcal{K}$ is a compact and convex subset of $M_2(\mathbb{R})$. b) Show that there exists $A \in \mathcal{K}$ such that $\det A = \sup_{B \in \mathcal{K}} \det B$.

We fix an element $A$ of $\mathcal{K}$ such that $\det A = \sup_{B \in \mathcal{K}} \det B$. Show that $\det A > 0$ and that there exists $x \in \mathcal{C}$ such that $\|Ax\| = 1$.

We define $Q_{>0} = (\mathbb{R}_+^*)^{I \times J}$ and $\mathscr{L} : Q_{>0} \times (\mathbb{R}^I \times \mathbb{R}^J) \rightarrow \mathbb{R}$ defined by $$\mathscr{L}(\boldsymbol{q}, (f, g)) = J_\epsilon(\boldsymbol{q}) + \sum_{i \in I} f_i \left(\alpha_i - \sum_{j \in J} q_{ij}\right) + \sum_{j \in J} g_j \left(\beta_j - \sum_{i \in I} q_{ij}\right).$$ (a) Show that for all $(f, g) \in \mathbb{R}^I \times \mathbb{R}^J$, the minimum of $\boldsymbol{q} \mapsto \mathscr{L}(\boldsymbol{q}, (f, g))$ on $Q_{>0}$ is attained at $q(f,g)_{ij} = e^{(f_i + g_j - C_{ij})/\epsilon} p_{ij}$.
(b) Calculate the value of $G(f, g) = \mathscr{L}(q(f,g), (f,g))$.
(c) Verify that $G$ is concave on $\mathbb{R}^I \times \mathbb{R}^J$.

Verify that if $f_* : \mathbb{R}^J \rightarrow \mathbb{R}^I$ and $g_* : \mathbb{R}^I \rightarrow \mathbb{R}^J$ are defined by $$f_*(g)_i = -\epsilon \log\left(\sum_{j \in J} e^{(g_j - C_{ij})/\epsilon} \beta_j\right) \text{ and } g_*(f)_j = -\epsilon \log\left(\sum_{i \in I} e^{(f_i - C_{ij})/\epsilon} \alpha_i\right)$$ then for all $(f, g) \in \mathbb{R}^I \times \mathbb{R}^J$, we have $\frac{\partial G}{\partial f_i}(f_*(g), g) = \frac{\partial G}{\partial g_j}(f, g_*(f)) = 0$ for all $(i,j) \in I \times J$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
We admit that $f$ and $g _ { s }$ are of class $\mathcal { C } ^ { 1 }$ on $U _ { n }$. Give the expression of their gradient at a point $x = \left( x _ { 1 } , \ldots , x _ { n } \right)$ of $U _ { n }$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
Prove that the restriction of $f$ to $X _ { s }$ admits a maximum on $X _ { s }$ and that this maximum is in fact attained on $X _ { s } \cap U _ { n }$.
You may verify that $f$ is strictly positive at certain points of $X _ { s } \cap U _ { n }$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
We denote by $a = \left( a _ { 1 } , \ldots , a _ { n } \right)$ an element of $X _ { s } \cap U _ { n }$ at which the restriction of $f$ to $X _ { s }$ attains its maximum.
Prove that there exists a real number $\lambda > 0$ such that, for all $k$ in $\llbracket 1 , n \rrbracket , a _ { k } = \frac { f ( a ) } { \lambda }$.