A query regarding Functions on Real Numbers Given two functions: $F(a)=b$, $G(c)=d$.
Let $Max(|b|, |d|)= m$
where: $a, b, c, d, m$ are real numbers.
Consider a statement:  For any set of values $(a_0, b_0, c_0, d_0, m_0)$ for above functions there exists another set of values $(a_1, b_1, c_1, d_1, m_1)$ such that $m_1<m_0$. 
Does the above statement prove that there exists a value $(a_i, c_i)$ such that $b_i = d_i = 0$?
 A: Not necessarily. Consider the identical functions:
$F(x) = G(x) = 1/x$
For any $a_0$, $c_0$ both nonzero, we have $b_0 = 1/a_0$, $d_0 = 1/c_0$, with 
$$m_0 = \max(\left\lvert1/a_0\right\rvert,\left\lvert1/c_0\right\rvert) = 
\frac{1}{\min(\left\lvert a_0 \right\rvert,\left\lvert c_0 \right\rvert)}$$
Choose $a_1 = c_1 = \max(\left\lvert a_0 \right\rvert,\left\lvert c_0 \right\rvert) + 1 > \min(\left\lvert a_0 \right\rvert,\left\lvert c_0 \right\rvert)$
Then $m_1 = \max(\left\lvert 1/a_1 \right\rvert, \left\lvert 1/c_1 \right\rvert) = 1/a_1 < 1/\min(\left\lvert a_0 \right\rvert,\left\lvert c_0 \right\rvert) = m_0$.
Thus this set of values $(a_1,b_1,c_1,d_1,m_1)$ satisfies the requirements.
However $F(x), G(x)$ never takes on the value $0$, so we can't have any $b_i = d_i = 0$.
EDIT: The statement cannot be true. For all sets $(a_i, b_i, c_i, d_i, m_i), m_i$ must be nonnegative. If there was some $b_i = d_i = 0$, then $m_i = \max(b_i, d_i) = 0$. 
However then the statement "For any set of values $(a_0, b_0, c_0, d_0, m_0)$, for above functions there exists another set of values $(a_1, b_1, c_1, d_1, m_1)$ s.t. $m_1 < m_0$" implies there is some $m_j$ s.t. $m_j < 0$, which is a contradiction.
