# monotonicity proof - double induction

Suppose we have 4 functions depending on a variable $x$ : $a(x), b(x), c(x)$ and $d(x)$.

I want to prove that $a(x) - c(x)$ is monotonically increasing in $x$, by an induction argument. Also I want to as well prove the same result for $b(x) - d(x)$.

In my derivations I obtain the following equation:

$a(x+1) - c(x+1)$ = $constant_1$ *$(\min(a(x), b(x)) - \min(c(x), d(x)))$

also

$b(x+1) - d(x+1)$ = $constant_2$ *$(\min(a(x), b(x)) - \min(c(x), d(x)))$

My question is if I claim that $a(x) - c(x)$ is increasing in x, also $b(x) - c(x)$ is increasing in $x$, then $(\min(a(x), b(x)) - \min(c(x), d(x)))$ is increasing in $x$?

I am not sure about this type of induction argument as it requires 2 conditions to happen at once for it to be true. Could that be an example of double induction though?

Thanks

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please use LateX formatting. – nbubis Mar 5 '13 at 22:50
The proof does not work unless you can show more abot $a,b,c,d$. For example, even if $c_1=1000$, then you might have $a(x)=10000, b(x)=1, c(x)=10, d(x)=0$, hence $a(x+1)-c(x+1)=1000\cdot(1-0)=1000 < 9990 = a(x)-c(x)$. – Hagen von Eitzen Mar 5 '13 at 22:55
@HagenvonEitzen Thanks. I only know in addition that all the functions are individually increasing in $x$ but that does not help either. – Roark Mar 5 '13 at 23:04
@HagenvonEitzen With the same setting I can affirm the following though: If $a(x) - c(x)$ non-decreasing in $x$ and $b(x) - c(x)$ non-decreasing in $x$ then $min(a(x), b(x)) - min(c(x), d(x))$ is also non-decreasing in $x$. Correct? – Roark Mar 6 '13 at 0:09