Show that if $x$ is a real number different from 1 and from $-1$, then $x ^ { 2 } - 2 x \cos \theta + 1 > 0$ for all $\theta \in \mathbb { R }$.

We place ourselves in the particular case where $E = \mathbb{R}_{2m}[X]$, with $m \geq 2$ a fixed natural integer. This vector space is equipped with the scalar product $$\forall (P,Q) \in E^2, \quad (P \mid Q) = \int_{-1}^{1} P(t)Q(t)\,dt$$ We set $$\mathbb{R}_{2m-1}^0[X] = \{P \in \mathbb{R}_{2m-1}[X] \mid P(-1) = 0 \text{ and } P(1) = 0\}$$
Show that $$\forall P \in \mathbb{R}_{2m-1}^0[X], \quad (P \mid P) \leq 4 (P' \mid P')$$ with strict inequality if $P$ is non-zero.

Show that $\lambda ^ { \alpha } \leq 1 + \lambda$ for all real $\lambda > 0$ and deduce the inequality:
$$\left| x - \frac { k } { n } \right| ^ { \alpha } \leq n ^ { - \alpha / 2 } \left( 1 + \sqrt { n } \left| x - \frac { k } { n } \right| \right)$$
for all $x \in ]0,1[ , n \in \mathbb { N } ^ { * }$ and $k \in \{ 0 , \ldots , n \}$.

Let $X : \Omega \rightarrow \mathbb{R}$ be a real-valued random variable. We assume that there exist two strictly positive reals $a$ and $b$ such that, for all non-negative reals $t$,
$$\mathbb{P}(|X| \geqslant t) \leqslant a \exp(-bt^{2})$$
Let $\delta$ be a real such that $0 \leqslant |\delta| \leqslant \sqrt{\frac{a}{b}}$. Show that, for all reals $t$,
$$-b(t - |\delta|)^{2} \leqslant a - \frac{1}{2}bt^{2}$$

We consider a convex function $f \in \mathcal{C}(\mathbb{R})$, admitting a minimizer $x_* \in \mathbb{R}$, and $\tau > 0$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := p_f(x_n)$. Show that for all $M, N \in \mathbb{N}$ $$|x_N - x_M| \leq \sqrt{2\tau|N-M|}\sqrt{\left|f(x_M) - f(x_N)\right|}$$

If $a, b, c, d$ are real numbers such that $a - b^2 \geq 1/4$, $b - c^2 \geq 1/4$, $c - d^2 \geq 1/4$, $d - a^2 \geq 1/4$, find the number of solutions $(a, b, c, d)$.
(A) 0 (B) 1 (C) 2 (D) Infinitely many

15. If $\mathrm { a } , \mathrm { b } , \mathrm { c }$ are positive real numbers, then prove that $[ ( 1 + \mathrm { a } ) ( 1 + \mathrm { b } ) ( 1 + \mathrm { c } ) ] ^ { 7 } > 7 ^ { 7 } \mathrm { a } ^ { 4 } \mathrm {~b} ^ { 4 } \mathrm { c } ^ { 4 }$.
Sol. $( 1 + a ) ( 1 + b ) ( 1 + c ) = 1 + a b + a + b + c + a b c + a c + b c$ $\Rightarrow \frac { ( 1 + \mathrm { a } ) ( 1 + \mathrm { b } ) ( 1 + \mathrm { c } ) - 1 } { 7 } \geq ( \mathrm { ab } . \mathrm { a } . \mathrm { b } . \mathrm { c } . \mathrm { abc } . \mathrm { ac } . \mathrm { bc } ) ^ { 1 / 7 } \quad ($ using $A M \geq G M )$ $\Rightarrow ( 1 + \mathrm { a } ) ( 1 + \mathrm { b } ) ( 1 + \mathrm { c } ) - 1 > 7 \left( \mathrm { a } ^ { 4 } \cdot \mathrm {~b} ^ { 4 } \cdot \mathrm { c } ^ { 4 } \right) ^ { 1 / 7 }$ $\Rightarrow ( 1 + \mathrm { a } ) ( 1 + \mathrm { b } ) ( 1 + \mathrm { c } ) > 7 \left( \mathrm { a } ^ { 4 } \cdot \mathrm {~b} ^ { 4 } \cdot \mathrm { c } ^ { 4 } \right) ^ { 1 / 7 }$ $\Rightarrow ( 1 + \mathrm { a } ) ^ { 7 } ( 1 + \mathrm { b } ) ^ { 7 } ( 1 + \mathrm { c } ) ^ { 7 } > 7 ^ { 7 } \left( \mathrm { a } ^ { 4 } \cdot \mathrm {~b} ^ { 4 } \cdot \mathrm { c } ^ { 4 } \right)$.

17. Prove that $\sin \mathrm { x } + 2 \mathrm { x } \geq \frac { 3 \mathrm { x } \cdot ( \mathrm { x } + 1 ) } { \pi } \forall \mathrm { x } \in \left[ 0 , \frac { \pi } { 2 } \right]$. (Justify the inequality, if any used).
Sol. Let $\mathrm { f } ( \mathrm { x } ) = 3 \mathrm { x } ^ { 2 } + ( 3 - 2 \pi ) \mathrm { x } - \pi \sin \mathrm { x }$ $\mathrm { f } ( 0 ) = 0 , \mathrm { f } \left( \frac { \pi } { 2 } \right) = - \mathrm { ve }$ $\mathrm { f } ^ { \prime } ( \mathrm { x } ) = 6 \mathrm { x } + 3 - 2 \pi - \pi \cos \mathrm { x }$ $\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) = 6 + \pi \sin \mathrm { x } > 0$ $\Rightarrow \mathrm { f } ^ { \prime } ( \mathrm { x } )$ is increasing function in $\left[ 0 , \frac { \pi } { 2 } \right]$ ⇒ there is no local maxima of $\mathrm { f } ( \mathrm { x } )$ in $\left[ 0 , \frac { \pi } { 2 } \right]$ ⇒ graph of $\mathrm { f } ( \mathrm { x } )$ always lies below the x -axis [Figure] in $\left[ 0 , \frac { \pi } { 2 } \right]$. $\Rightarrow \mathrm { f } ( \mathrm { x } ) \leq 0$ in $\mathrm { x } \in \left[ 0 , \frac { \pi } { 2 } \right]$. $3 \mathrm { x } ^ { 2 } + 3 \mathrm { x } \leq 2 \pi \mathrm { x } + \pi \sin \mathrm { x } \Rightarrow \sin \mathrm { x } + 2 \mathrm { x } \geq \frac { 3 \mathrm { x } ( \mathrm { x } + 1 ) } { \pi }$.