Maximum number of positive roots is $3$ Let $$f(x) = a+b(x+p)^t+c(x+p)^t(x+q)^t+d(x+p)^t(x+q)^t(x+r)^t,$$
where $t>1$ is any positive real number, $p>q>r>0$  or  $p<q<r$ are positive integers and $a,b,c,d$ are any arbitrary real numbers which are not simultaneously zero.  
Que: Prove that,
$1.$ $f(x)$ has at most $3$ positive zeros. 
$2.$ $f(x)$ has at most $3$ positive integer zeros. 
I need $2$ actually, but I think $1$ is also true. 
My attempts: 
I tried to show that $f'(x)$ has at most two positive zeros so that by Rolle's Theorem, the result follows, but the expression for $f'(x)$ is not fruitful.
Note: This question is also on Mathstackexchange, (https://math.stackexchange.com/questions/4155919/does-this-function-have-at-most-3-positive-zeros). 
Kindly suggest any help or tip. Thanks in advance.
 A: 
Claim 1: $f$ has $\le3$ positive roots if $t\ge1$ and $p,q,r$ are integers such that $0<p<q<r$.


Claim 2: $f$ has $\le3$ positive roots if $t\ge2$ and $p,q,r$ are integers such that $p>q>r>0$.

Let
$$f_1(x):=\frac{f'(x)}{(p+x)^{t-1}},$$
$$f_2(x):=\frac{f_1'(x)}{(q+x)^{t-2}(q + p (t-1) + q t + 2 t x)},$$
$$f_3(x):=\frac{f_2'(x)}{t(r+x)^{t-3}}\,(q + p (t-1) + q t + 2 t x)^2.$$
If $d=0$, then $f_2(x)=ct$. So, if $c\ne0$, then $f_2$ has no positive roots, and hence $f_1$ has $\le1$ positive roots, so that $f$ has $\le2\le3$ positive roots. If $d=0=c$, then $f(x)=a+b(x+p)^t$, so that $f$ has $\le1\le3$ positive roots, because $|a|+|b|\ne0$.
It remains to consider the case $d\ne0$.
Consider the case when $t\ge1$ and $p,q,r$ are positive integers such that $0<p<q<r$, as in Claim 1.
Then, letting $S:=t-1[\ge0]$, $P:=p-1[\ge0]$, $Q:=q-p-1[\ge0]$, and $R:=r-q-1[\ge0]$, we see that $f_3(x)/d$ is a polynomial in $x,S,P,Q,R$ with all coefficients nonnegative and a positive free term. So, $f_3$ has no positive roots, and hence $f_2$ has $\le1$ positive roots, and hence $f_1$ has $\le2$ positive roots, so that $f$ has $\le3$ positive roots. This completes the proof of Claim 1.
Consider now the case when $t\ge2$ and $p,q,r$ are positive integers such that $p>q>r>0$, as in Claim 2.
Then, letting $S:=t-2[\ge0]$, $P:=p-q-1[\ge0]$, $Q:=q-r-1[\ge0]$, and  $R:=r-1[\ge0]$, we see that $f_3(x)/d$ is a polynomial in $x,S,P,Q,R$ with all coefficients nonnegative. So, $f_3$ has no positive roots, and hence $f_2$ has $\le1$ positive roots, and hence $f_1$ has $\le2$ positive roots, so that $f$ has $\le3$ positive roots. This completes the proof of Claim 2.

Here are details of the calculations, with Mathematica (click on the images to enlarge them):


