Proof by induction in inequalities $$\sum^n_{k=1} \frac1{k^3} \le \frac 5 4 - \frac 1 {2n^2}$$ For all $n\ge 2$
Now this really a tough one for me.
The base case holds at $n = 2$
Then i replaced it with $p$ and then $p+1$
I got an inequality where i deduced that $p+1 = p$ and then plus a 1. But its not burging
 A: Assume that the inequality be equal for $n=p$ as you did
$$
\sum\limits_{k = 1}^p {{1 \over {k^{\,3} }}}  \le {5 \over 4} - {1 \over {2p^{\,2} }}
$$
Then it ia also valid for $n=p+1$
$$
\sum\limits_{k = 1}^{p + 1} {{1 \over {k^{\,3} }}}  = \sum\limits_{k = 1}^p {{1 \over {k^{\,3} }}}  + {1 \over {\left( {p + 1} \right)^{\,3} }}
 \le \left( {{5 \over 4} - {1 \over {2p^{\,2} }}} \right) + {1 \over {\left( {p + 1} \right)^{\,3} }} \le {5 \over 4} - {1 \over {2\left( {p + 1} \right)^{\,2} }}
$$
because in fact, we can write in sequence
$$
\eqalign{
  & \left( {{5 \over 4} - {1 \over {2p^{\,2} }}} \right) + {1 \over {\left( {p + 1} \right)^{\,3} }} \le {5 \over 4} - {1 \over {2\left( {p + 1} \right)^{\,2} }}  \cr 
  &  - {1 \over {2p^{\,2} }} + {1 \over {\left( {p + 1} \right)^{\,3} }} \le  - {1 \over {2\left( {p + 1} \right)^{\,2} }}  \cr 
  &  - {{\left( {p + 1} \right)^{\,2} } \over {p^{\,2} }} + {2 \over {\left( {p + 1} \right)}} \le  - 1  \cr 
  &  - \left( {p + 1} \right)^{\,3}  + 2p^{\,2}  \le  - p^{\,2} \left( {p + 1} \right)  \cr 
  & p^{\,2} \left( {p + 1} \right) \le \left( {p + 1} \right)^{\,3}  - 2p^{\,2}   \cr 
  & p^{\,3}  + 3p^{\,2}  \le \left( {p + 1} \right)^{\,3}  = p^{\,3}  + 3p^{\,2}  + 3p + 1 \cr} 
$$
A: HINT:
Induction hypothesis: Suppose that $\sum^n_{k=1} \frac1{k^3} \le \frac 5 4 - \frac 1 {2n^2}$ holds for some $n$.
Consider $n+1$: $$\sum^{n+1}_{k=1} \frac1{k^3}\le\frac54+\frac1{(n+1)^3}-\frac1{2n^2}$$ and \begin{align}\frac1{(n+1)^3}-\frac1{2n^2}\le\frac1{2(n+1)^2}&\impliedby 2n^2-(n+1)^3\le n^2\\&\impliedby\cdots\end{align}
