Simultaneous version of Triangle inequality

Theorem 1. If $\alpha_1,\dots,\alpha_m$ are arbitrary complex numbers, $m\geq2$, and if $H\in\mathbb{N}$, then there exist numbers 
\begin{align} a_k\in\mathbb{Z}, \quad \lvert a_k\rvert\leq H, \quad k=1,\dots,m, \quad \max_{1\leq k\leq m} \lvert a_k\rvert>0 \end{align}
such that the linear form 
\begin{align} L=a_1\alpha_1+\cdots+a_m\alpha_m \end{align}
satisfies the inequality 
\begin{align} \lvert L\rvert\leq cH^{1-\tau m} \end{align}
where $\tau=1$ if all of the $\alpha_k$ are real and $\tau=\frac{1}{2}$ if one or more of them is complex, and
$$c=\left\{ \begin{array}{ll} \sum_{k=1}^m \lvert \alpha_k\rvert, & \quad \text{if}\;\tau=1 \\ \sqrt{2}\sum_{k=1}^m \lvert \alpha_k\rvert, & \quad \text{if}\;\tau=\frac{1}{2} \end{array} \right.$$

Proof of Theorem 1. The theorem is obvious if $H=1$ by the generalization of Triangle inequality; so we suppose that $H\geq 2$. We consider all possible linear forms $L$, where the $a_k$ independently run through all integer values satisfying the inequalities
\begin{align} \lvert a_k\rvert\leq \bigg[\frac{H}{2}\bigg], \quad k=1,\dots,m\notag \end{align}
The number of such forms $L$ is equal to 
\begin{align} \bigg(2\bigg[\frac{H}{2}\bigg]+1\bigg)^m \end{align}
Their values satisfy the inequalities 
\begin{align} \lvert L\rvert\leq \gamma\bigg[\frac{H}{2}\bigg], \quad \gamma=\sum_{k=1}^m \lvert\alpha_k\rvert \end{align}
We consider two cases (we exclude the obvious case $\gamma=0$).
    1. All of the $\alpha_1,\dots,\alpha_m$ are real numbers. Then the values of all of our forms $L$ are contained in the interval with endpoints at $\pm\gamma[\frac{H}{2}]$, which has length $2\gamma[\frac{H}{2}]$. We divide this interval into
\begin{align} \bigg(2\bigg[\frac{H}{2}\bigg]+1\bigg)^m-1    \end{align}
equal subintervals. The number of forms $L$ is greater than the number of subintervals. Hence, there exists a subinterval containing the values of two different forms $L$. Let these two forms be
\begin{align} L'=a_1'\alpha_1+\cdots+a_m'\alpha_m, \quad L''=a_1''\alpha_1+\cdots+a_m''\alpha_m \end{align}
Then
$$\lvert L'-L''\rvert\leq \frac{2\gamma \Big[\frac{H}{2}\Big]}{\Big(2\Big[\frac{H}{2}\Big]+1\Big)^m-1}$$
If $H$ is an even number, then $2[\frac{H}{2}]=H$, and since $(H+1)^m>H^m+1$, we have
$$\lvert L'-L''\rvert\leq \frac{\gamma H}{(H+1)^m-1}<\gamma H^{1-m}$$
If $H$ is an odd number, then $2[\frac{H}{2}]=H-1$, and since $H^{1-m}<1$ for $m\geq2$, we have
$$\lvert L'-L''\rvert\leq \frac{\gamma (H-1)}{H^m-1}<\gamma H^{1-m}$$
Thus
$$\lvert L'-L''\rvert<\gamma H^{1-m}$$
We set $a_k=a_k'-a_k''$, $k=1,\dots,m$. Then the form 
$$L=L'-L''=a_1\alpha_1+\cdots+a_m\alpha_m$$
satisfies the conditions
$$\lvert a_k\rvert\leq 2\bigg[\frac{H}{2}\bigg]\leq H, \quad \sum_{k=1}^m \lvert a_k\rvert>0$$
and
$$\lvert L\rvert<cH^{1-m}, \quad c=\gamma$$
    2. At least one of the numbers $\alpha_1,\dots,\alpha_m$ is complex. In this case it follows from $\lvert L\rvert\leq \gamma[\frac{H}{2}]$ that the values of the forms $L$ lie in a square centered at the origin with sides parallel to the coordinate axes of length equal to $2\gamma[\frac{H}{2}]$. We divide the sides of this square into $M$ equal segments, where $M$ is the integer satisfying the inequalities
\begin{align} \bigg(2\bigg[\frac{H}{2}\bigg]+1\bigg)^{\frac{m}{2}}-1\leq M< \bigg(2\bigg[\frac{H}{2}\bigg]+1\bigg)^{\frac{m}{2}}   \end{align}
Using lines through the division points parallel to the coordinate axes, we divide our square into $M^2$ small squares. From the inequality above we have
$$M^2<\bigg(2\bigg[\frac{H}{2}\bigg]+1\bigg)^{m}$$
This inequality means that the number of small squares is less than the number of linear forms $L$. Hence, there exists a small square which contains the values of two different forms $L$. Suppose that these two forms are the ones in case 1. We then find that $\lvert L'-L''\rvert$ is no greater than the diagonal of a small square. This observation, together with the inequalities, enables us to say that the following inequalities hold for any $H\geq2$ and $m\geq2$:
\begin{align} \lvert L'-L''\rvert\leq \frac{\sqrt{2}\gamma2\Big[\frac{H}{2}\Big]}{M}\leq \frac{\sqrt{2}\gamma2\Big[\frac{H}{2}\Big]}{\Big(2\Big[\frac{H}{2}\Big]+1\Big)^{\frac{m}{2}}-1}   \end{align}
If $H$ is an even number, then $2[\frac{H}{2}]=H$, and since $m\geq2$, we conclude that
$$\lvert L'-L''\rvert\leq \frac{\sqrt{2}\gamma H}{(H+1)^{\frac{m}{2}}-1}\leq \sqrt{2}\gamma H^{1-\frac{m}{2}}$$
If $H$ is an odd number, then $2[\frac{H}{2}]=H-1$, and we similarly conclude that
$$\lvert L'-L''\rvert\leq \frac{\sqrt{2}\gamma (H-1)}{H^{\frac{m}{2}}-1}=\sqrt{2}\gamma H^{1-\frac{m}{2}}\frac{1-H^{-1}}{1-H^{-\frac{m}{2}}}\leq \sqrt{2}\gamma H^{1-\frac{m}{2}}$$
We then argue as in the first case in order to obtain the inequality $\lvert L\rvert \leq cH^{1-\frac{m}{2}}$ with $c=\sqrt{2}\gamma$. $\Box$

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