How to prove there exist integer $u,v\in Z$,such $|f(u,v)|\le\sqrt{\frac{4D}{3}}$?

let $a,b,c\in R$,and such $D=ac-b^2>0$,and $$f(x,y)=ax^2+2bxy+cy^2$$,show that:There exist integer $u,v\in Z((u,v)\neq (0,0)$),such $$|f(u,v)|\le\sqrt{\dfrac{4D}{3}}$$

Question 1: How to prove this result?

Qustion 2: This looks like a very interesting result. Is this an existing result(It seems to have a geometric meaning)? If yes,

please tell me the background or Which book deals with this issue in detail? Thank you

I try:since $ac-b^2>0$,then $ac>b^2\ge 0$ so have two case $a<0,c<0$ or $a>0,c>0$

(1):if $a<0,c<0$,since $D=ac-b^2>0$ and $$f(x,y)=ax^2+2bxy+cy^2=a\left(x+\frac{b}{a}y\right)^2+\dfrac{ac-b^2}{a}y^2<0$$ so this case it clear exist integer $u,v$ such $$f(u,v)<0<\sqrt{\frac{4D}{3}}$$

(2): But $a>0,c>0$ this case $f(x,y)>0$ But How to prove $f(u,v)<\sqrt{\frac{4D}{3}}$

Let $\alpha,\overline\alpha$ be the two roots of the polynomial $t^2+\frac{2b}{a}t+\frac{c}{a}$. By our assumption, its discriminant is negative, so $\alpha$ is a complex number. For any $x,y\in\mathbb Z$ we have $$ax^2+2bxy+cy^2=ay^2\left(\frac{x}{y}-\alpha\right)\left(\frac{x}{y}-\overline\alpha\right)=|\sqrt{a}x-\sqrt{a}\alpha y|^2,$$ so our question is reduced to showing that if we consider a lattice in $\mathbb C$ spanned by vectors $\sqrt{a}$ and $\sqrt{a}\alpha$, then it contains a vector of length at most $\sqrt[4]{\frac{4D}{3}}$. After we realize that $D$ coincides with the discriminant of that lattice, we have reduced the following general fact about lattices:

Any lattice in $\mathbb R^2$ of discriminant $D$ contains a vector of length at most $\sqrt[4]{\frac{4D}{3}}$.

This is essentially a restatement of the fact that the second Hermite constant is equal to $\gamma_2=\frac{2}{\sqrt{3}}$.

For completeness, here is an easy proof of this fact. After some rescaling, the statement is equivalent to: any lattice in $\mathbb R^2$ such that $(1,0)$ is its shortest vector has covolume at least $\sqrt{\frac{3}{4}}$. Indeed, there is a basis of the lattice of the form $\{(1,0),(x,y)\}$, and then its covolume is equal to $y$. Translating $(x,y)$ by a multiple of $(1,0)$, we can assume $|x|\leq\frac{1}{2}$. Since $(1,0)$ is the shortest vector, we need to have $x^2+y^2\geq 1$, hence $y^2\geq\frac{3}{4}$, so $y\geq\sqrt{\frac{3}{4}}$.