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 $x_0 \in \bar{U}$.

Determine $\max_{1 \leqslant p \leqslant n-1} \left( n^{2} - pn + p^{2} \right)$.

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 $f''(x_0) < 0$.

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 $f ^ { \prime \prime } \left( x _ { 0 } \right) < 0$.

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 $\mathbb{R}^I \times \mathbb{R}^J$.

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 $\det A = \sup_{B \in \mathcal{K}} \det B$.

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 $(i,j) \in I \times J$.

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 $\|Ax\| = 1$.

Let $a \in E$ and let $U$ be an open set of $E$ containing $a$. Let $f : U \rightarrow E$ be an application of class $\mathscr{C}^1$ on $U$ such that $df(a) = \operatorname{Id}_E$. We fix a real number $r > 0$ such that $\overline{B(a,r)} \subset U$ and $$\forall x_1, x_2 \in \overline{B(a,r)}, \quad \left\|f(x_1) - f(x_2)\right\| \geqslant \frac{1}{2} \left\|x_1 - x_2\right\|.$$ Let $y_0 \in E$ such that $\left\|y_0 - f(a)\right\| \leqslant \frac{r}{4}$.
Show that the application $$\begin{array}{ccc} \overline{B(a,r)} & \longrightarrow & \mathbb{R} \\ x & \longmapsto & \left\|y_0 - f(x)\right\|^2 \end{array}$$ admits a minimum attained at a point $x_0$ of $B(a,r)$.

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 $\llbracket 1 , n \rrbracket , a _ { k } = \frac { f ( a ) } { \lambda }$.

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 $\left( \mathbb { R } _ { + } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $h _ { n }$ the map from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad h _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \cdots + x _ { n } - 1 .$$
We denote by $H _ { n }$ the set $H _ { n } = \left\{ \left( x _ { 1 } , \ldots , x _ { n } \right) \in \mathbb { R } ^ { n } \mid x _ { 1 } + \cdots + x _ { n } = 1 \right\}$. We denote by $M _ { n }$ the maximum of $F _ { n }$ on $\overline { U _ { n } } \cap H _ { n }$ and we denote by $( a _ { 1 } , \ldots , a _ { n } )$ a point of $U _ { n } \cap H _ { n }$ at which it is attained. For $k$ between 1 and $n$, we denote $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$.
Prove that there exists a real number $\lambda > 0$ such that
$$\left\{ \begin{aligned} \gamma _ { 1 } + \frac { \gamma _ { 2 } } { 2 } + \cdots + \frac { \gamma _ { n } } { n } & = \lambda a _ { 1 } \\ \frac { \gamma _ { 2 } } { 2 } + \cdots + \frac { \gamma _ { n } } { n } & = \lambda a _ { 2 } \\ & \vdots \\ \frac { \gamma _ { n } } { n } & = \lambda a _ { n } \\ a _ { 1 } + a _ { 2 } + \cdots + a _ { n } & = 1 \end{aligned} \right.$$

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 $\left( \mathbb { R } _ { + } \right) ^ { n }$. We consider the map $F _ { n }$ from $\overline { U _ { n } }$ to $\mathbb { R }$, defined by
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } } , \quad F _ { n } \left( x _ { 1 } , \ldots , x _ { n } \right) = x _ { 1 } + \left( x _ { 1 } x _ { 2 } \right) ^ { 1 / 2 } + \left( x _ { 1 } x _ { 2 } x _ { 3 } \right) ^ { 1 / 3 } + \cdots + \left( x _ { 1 } \cdots x _ { n } \right) ^ { 1 / n } .$$
We denote by $M _ { n }$ the maximum of $F _ { n }$ on $\overline { U _ { n } } \cap H _ { n }$ and we denote by $( a _ { 1 } , \ldots , a _ { n } )$ a point of $U _ { n } \cap H _ { n }$ at which it is attained. For $k$ between 1 and $n$, we denote $\gamma _ { k } = \left( a _ { 1 } a _ { 2 } \cdots a _ { k } \right) ^ { 1 / k }$. There exists a real number $\lambda > 0$ satisfying the system in Q17.
Deduce that:
a) $\lambda = \gamma _ { 1 } + \gamma _ { 2 } + \cdots + \gamma _ { n } = M _ { n }$;
b) for all $k$ in $\llbracket 1 , n \rrbracket , \gamma _ { k } = \lambda \omega _ { k } a _ { k }$, where
$$\left\{ \begin{array} { l } \omega _ { k } = k \left( 1 - \frac { a _ { k + 1 } } { a _ { k } } \right) \text { if } k \in \llbracket 1 , n - 1 \rrbracket \\ \omega _ { n } = n \end{array} \right.$$

Consider the function $$f(x) := \frac{1}{3}x^3 \quad \text{if } x \geq 0, \quad f(x) := 0 \quad \text{if } x < 0$$ Justify that $f \in \mathcal{C}^1(\mathbb{R})$ and that $f$ is convex. Give the set of its minimizers.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. a) Show that for all $x, y \in \mathbb{R}$ $$f(y) \geq f(x) + f'(x)(y-x)$$ Hint: consider a Taylor expansion of the convexity inequality when $t \rightarrow 0^+$. b) Show that for all $x, y \in \mathbb{R}$ $$f(y) \leq f(x) + f'(x)(y-x) + \frac{L}{2}(y-x)^2$$ c) Establish that for all $n \in \mathbb{N}$ $$f(x_{n+1}) \leq f(x_n) - \frac{\tau}{2}(2 - \tau L)\left|f'(x_n)\right|^2$$ Deduce that the sequence $\left(f(x_n)\right)_{n \in \mathbb{N}}$ is decreasing.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. Establish an upper bound for the sequence with general term $a_n := f(x_n) - f(x_*)$. Conclude that $f(x_n) \rightarrow f(x_*)$ when $n \rightarrow \infty$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. Show that $\frac{\tau}{2}(2 - \tau L)\sum_{0 \leq i < n}\left|f'(x_i)\right|^2 \leq \left(f(x_0) - f(x_n)\right)$ for all $n \in \mathbb{N}^*$. Deduce that $f'(x_n) \rightarrow 0$ when $n \rightarrow \infty$.

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. a) Show that the sequence $\left(x_n\right)_{n \in \mathbb{N}}$ admits a convergent subsequence. We denote by $\varphi : \mathbb{N} \rightarrow \mathbb{N}$ the extractor and $x_{**}$ the corresponding limit, so that $x_{\varphi(n)} \rightarrow x_{**}$ when $n \rightarrow \infty$. Hint. We may use without proof the Bolzano-Weierstrass theorem: from any bounded sequence in $\mathbb{R}$, we can extract a convergent subsequence. b) Show that $f'(x_{**}) = 0$. c) Deduce that $x_{**}$ is a minimizer of $f$, then that $\left|x_n - x_{**}\right| \rightarrow 0$ when $n \rightarrow \infty$.

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$. Let also $\tau > 0$. Show that the function $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$ admits a unique minimizer on $\mathbb{R}$, which we will denote $p_f(x_0)$. Hint: We may consider minimizers $x_1$ and $x_2$ of $F_{x_0}$, and note that $$\left|\frac{1}{2}(x_1 + x_2) - x_0\right|^2 < \frac{1}{2}|x_1 - x_0|^2 + \frac{1}{2}|x_2 - x_0|^2 \quad \text{if } x_1 \neq x_2$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The operator $p_f$ is defined as the unique minimizer of $F_{x_0}(x) := \frac{1}{2}|x - x_0|^2 + \tau f(x)$. Show that $x_0 \in \mathbb{R}$ is a minimizer of $f$ if and only if $p_f(x_0) = x_0$. Hint: consider the quantity $F_{x_0}((1-t)x_0 + tx_*)$ when $t \rightarrow 0^+$.

In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II.
We set $G_h : x \longmapsto \frac{(x-h)^2}{2\beta} - \ln(2\operatorname{ch}(x))$.
By distinguishing cases according to whether $\beta \leqslant 1$ or $\beta > 1$, show that $G_h$ admits a minimum on $\mathbb{R}_+$ attained at a unique point $u_h$. Show moreover that

• [(a)] if $\beta \leqslant 1$ and $h = 0$, then $u_h = 0$;
• [(b)] if $\beta > 1$, then $u_h > 0$;
• [(c)] for any $\beta \in \mathbb{R}_+^*$,
  • [(i)] $G_h'(u_h) = 0$;
  • [(ii)] $h = \beta G_0'(u_h)$;
  • [(iii)] $G_h''(u_h) > 0$ when $h > 0$.

In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II.
We set $G_h : x \longmapsto \frac{(x-h)^2}{2\beta} - \ln(2\operatorname{ch}(x))$.
We now assume that $h > 0$. Justify that $G_h$ attains its minimum on $\mathbb{R}$ at a unique point which is $u_h$.

Show that
$$\forall ( x , y ) \in ] 0,1 \left[ ^ { 2 } , \quad \frac { x ( 1 - x ) y ( 1 - y ) } { 1 - x y } \leqslant \frac { 5 \sqrt { 5 } - 11 } { 2 } \right.$$

Let $a$, $b$, $h$ be the three edges meeting at a particular vertex of a triangular prism, such that $a$, $b$ are sides of a base triangle with angle $\theta$ between them and $h$ is the height of the prism. Given that the total surface area is $K$, show that the volume $V$ satisfies $V \leq \sqrt{K^3/54}$, and find the dimensions of the prism of maximum volume.