Proving $\sum_{i=1}^{n}\frac{1}{p^{a_i}} \geq \sum_{i=1}^{n}\frac{1}{p^{b_i}}$, when $p \geq 2$. Suppose we are given $\sum_{i=1}^{n}\frac{1}{a_i} \geq  \sum_{i=1}^{n}\frac{1}{b_i}$, where $a_i,b_i >0$, $\forall i$. Is  $\sum_{i=1}^{n}\frac{1}{p^{a_i}} \geq  \sum_{i=1}^{n}\frac{1}{p^{b_i}}$, where $p \geq 2$?
The above seems to be intuitively true, but I haven't found a rigorous way to prove (or disprove) the above result. Any suggestions or pointers to relevant inequalities would be much appreciated. 
 A: No, it's not necessarily true. For example, let $n = 2$, $a_1 = 0.5$, $a_2 = 100$, $b_1 = b_2 = 1$ and $p = 2$. In this case,
$$\sum_{i = 1}^{n} \frac{1}{a_i} = 2 + .01 = 2.01 \tag{1}\label{eq1}$$
$$\sum_{i = 1}^{n} \frac{1}{b_i} = 1 + 1 = 2 \tag{2}\label{eq2}$$
Thus, your first condition holds. However,
$$\sum_{i = 1}^{n} \frac{1}{p^{a_i}} = \cfrac{1}{\sqrt{2}} + \cfrac{1}{2^{100}} < 1 \tag{3}\label{eq3}$$
but
$$\sum_{i = 1}^{n} \frac{1}{p^{b_i}} = \cfrac{1}{2} + \cfrac{1}{2} = 1 \tag{4}\label{eq4}$$
As such, your second condition doesn't hold. I don't believe there is any minimum value of $p$ you would allow your second condition to hold in all cases where the first condition holds, with the answer of Doyun Nam giving an example of how you can show this.
A: Your inequality isn't true in general case. 
Assume $n=3$, $a_1 = a_2 = a_3 = 2$, and $b_1=1, b_2=b_3 = 4$. Then your condition for $a_i$ and $b_i$ is satisfied. 
However, if your inequality is true, then 
$$\frac{3}{p^2} \geq \frac{1}{p} + \frac{2}{p^4}$$
should hold for all $p \geq 2$. 
But it doesn't, one of the counter example is the case $p=3$. 

It is another approach.
If we allow an extented real number system, let $n=2$, $a_1=a_2=2$, and $b_1=1, b_2 = \infty$. 
Then $\frac{1}{2} + \frac{1}{2} = \frac{1}{1} + \frac{1}{\infty}$. 
Thus the condition for $a_i$ and $b_i$ is satisfied. 
However, 
$$\frac{2}{p^2} \geq \frac{1}{p} + \frac{1}{p^\infty} = \frac{1}{p}$$
only holds where $2 \geq p$. 
