I would like help in this problem please. I don't write very well in English, because I'm Brazilian.
Consider the quadratic function $f(x) = \frac{1}{2}x^{\top}Ax + b^{\top}x$, with $A \in \mathbb{R}^{n\times n}$ symmetric and $b \in \mathbb{R}^n$. Show that if $f$ is bounded below, then $A$ is positive semidefinite and $f$ has a global minimizer.
What did I do: Suppose that $A$ is not positive semidefinite, that is, there is a nonzero vector $u \in \mathbb{R}^n $ such that $\langle Au, u\rangle< 0$. Therefore we have to prove that there is $k > 0$ such that $f(ku) < c$ be $k>0$ and $c \in R$ such that $$f(ku) = \langle A(ku), ku\rangle + \langle b, ku\rangle = k^{2}\langle Au,u\rangle + k\langle b,u\rangle< c.$$ I don't know if it's right and I'm not able to conclude
You are off to a good start; indeed if $A$ is not positive semidefinite then there exists some $u\in\Bbb{R}^{n\times n}$ such that $u^{\top}Au<0$. Let $C_0:=u^{\top}Au$ and $C_1:=b^{\top}u$ so that $f(u)=C_0+C_1$, and for every scalar $k\in\Bbb{R}$ you have $$f(ku)=(ku)^{\top}A(ku)+b^{\top}(ku)=k^2(u^{\top}Au)+k(b^{\top}u)=C_0k^2+C_1k.$$ Because $C_0<0$ it follows that $\lim_{k\to\infty}f(ku)=-\infty$, contradicting the assumption that $f$ is bounded below.