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The minimum here can be found exactly, in a finite number of steps. Indeed, the target function is concave. So, its minimum on the (compact) (transportation) polytope $P:=\Pi(\alpha,\beta)$ is attaine …

Section 1.3 of Chapter Nonsmooth Optimization by V. F. Demyanov in the book Nonlinear Optimization discusses in some detail the Dini, Hadamard, Shor, Clarke, and Michel–Penot notions of the subdiffere …

Let $c:=\Phi>0$ and $a:=t_{\max}>0$. For any given value (say $m$) of $ET$, the variance $VT$ of $T$ is maximized when $T$ takes only the endpoint values, $0$ and $a$, so that $P(T=a)=m/a=1-P(T=0)$, a …

This inequality is not true. E.g., $$g\Big(\frac1{10},\frac1{10},\frac\pi4\Big)=-1783.256\ldots\not\ge1.$$ The OP has changed the problem, thus invalidating the previous answer. After the change, the …

$\newcommand\R{\mathbb R}$ A convenient way to derive the dual problem from a primal one is by using the minimax duality for the Lagrangian, which is given here by the formula $$L(f,v):=\int_R\big[2| …

For $a\ge0$ and $u\ge0$, let $$q(u):=\ln Q(a+\sqrt u). $$ Then $$q_2(t):=q''(u)\frac{8 \sqrt{2 \pi } t^3 e^{\frac{1}{2} (a+t)^2} Q(a+t)^2}{a t+t^2+1} =2 Q(a+t)-\frac{\sqrt{\frac{2}{\pi }} t e^{-\fr …

Here is a method that will allow one to find the exact upper and lower bounds on $g(z)$ over $z>0$ with any degree of accuracy. Take any real $z>0$. Since \begin{equation*} \frac1y=\int_0^\infty d …

$\newcommand{\ga}{\gamma}$Letting $\ga\to\infty$ (with $A,B,\lambda,f(0)$ fixed), we see that the left-hand side of your inequality \begin{equation} \begin{aligned} &A + (\lambda + \gamma)^2 + \g …

Not in general. E.g., let $n=3$, $A(x_1,x_2,x_3):=1-x_1 x_2+2 x_3 x_2+x_1 x_3$, and $B(x_1,x_2,x_3):=2-x_1 x_2+x_3 x_2+2 x_1 x_3$. Then for $x_3\in(0,2\sqrt2)$ $$G(x_3)=\max_{t\in\mathbb R}\frac{A(t, …

Your question is Find the minimum of $f(x,y)=(x-2)^2+y$ subject to $y-x^3\ge0$, $y+x^3\le0$ and $y\ge0$. The restrictions $y-x^3\ge0$, $y+x^3\le0$, and $y\ge0$ can be rewritten as $0\le y\le-x^3$ or …

$\newcommand\R{\mathbb R}\newcommand\ep\epsilon\newcommand\bad{\text{bad}}$One would think that the answer is "of course no, descent does not have to monotonic". However, the no stationary points cond …

This ODE is extremely unlikely to have an explicit solution. Mathematica cannot do anything with this ODE even when $g(x)\equiv x$. Here is an image of the corresponding Mathematica notebook: And her …

Rewrite the inequality in question as \begin{equation*} f(u+v)\le f(u)+f(v) \end{equation*} for $u,v$ in $\mathbb R_+^4$, where \begin{equation*} f(u):=-\left(\left(\frac{1}{\sqrt{u_1}}+\frac{1} …

$\newcommand{\R}{\mathbb{R}}$ An advantage of my previous answer was that, while the computer calculations were pretty heavy there, the logic was extremely simple; virtually no thinking or ingenuity w …

Let $f\colon X\times Y\to\mathbb R$, where $X$ and $Y$ are any sets. Suppose that the function $f$ is generalized concave-convex in the sense that for any $x_0,x_1$ in $X$, any $y_0,y_1$ in $Y$, and a …