I was given the following statement:
$\sum_{k=1}^{n} p_k = 1$.
And I proved these two statements:
a) $\sum_{k=1}^{n} p_k^2 \geq \frac{1}{n}$ and
b) $\sum_{k=1}^{n} \frac{1}{p_k} \geq n^2$.
Is there a way I can extend the logic to the following statement?
$\sum_{k=1}^{n} \frac{1}{p_k^2} \geq n^3$
I assume I can't just take the inverse of statement $a$ and multiply it to statement $b$, and even if I did I would end up with $\sum_{k=1}^{n} \frac{1}{p_k^3} \geq n^3$, which is slightly different than what I'm trying to prove. If there isn't a way, I'll have to start chugging out a cauchy proof! Thank you for your insights.