How to prove the following inequality involving monotonic decreasing function? Question 1: If $0<\gamma_1<\gamma_2$, $N\in {\mathbb N}^{+}$,
$$\sum_{n=1}^{N} n^{-\frac{\gamma_1}{2}}\cdot \sqrt{\sum_{n=1}^{N}n^{-\gamma_2}}-\sum_{n=1}^{N} n^{-\frac{\gamma_2}{2}}\cdot \sqrt{\sum_{n=1}^{N}n^{-\gamma_1}}>0$$
Proving the above question is equivilent to proving the following question:
Question 2: The following equation below is a monotonic decreasing function of $\gamma$.
$$\forall N\in {\mathbb N}^{+}, f(\gamma)=\frac{\sum_{n=1}^{N} n^{-\frac{\gamma}{2}}}{\sqrt{\sum_{n=1}^{N}n^{-\gamma}}}$$
The following is the curve for $f(\gamma)$:

Can anyone prove the original question?
Added:
I have tried some methods to prove Q.1 or Q.2.
For Q.1, I tried to use some inequalities, but failed.
$$\text{e.g.} \ \ \sum_{n=1}^{N}x_n>\sqrt{\sum_{n=1}^{N}{x_n}^2}$$
For Q.2, I tried by proving $f'(\gamma)<0$. Similar problem as proving Q.1 will appear when proving $f'(\gamma)<0$. But, also failed.
Proving one of the two is OK.
 A: Write $f(\gamma) = \left( \sum_{n=1}^{N} n^{-\gamma/2} \right)\left( \sum_{n=1}^{N} n^{-\gamma} \right)^{-1/2}$. Then
$$
h(\gamma_) \equiv \log f(\gamma) = \log\left(\sum_{n=1}^{N} n^{-\gamma/2} \right) - \tfrac{1}{2} \log\left( \sum_{n=1}^{N} n^{-\gamma} \right)
$$
Now $\log(\cdot)$ is everywhere increasing so it is sufficient to show that $h(\gamma)$ is everywhere monotonic decreasing. We have
$$\begin{align*}
h'(\gamma)
&=
\frac{\sum_{n=1}^{N} -\tfrac{1}{2} n^{-\gamma/2} \log(n)}{\sum_{n=1}^{N} n^{-\gamma/2}}
-
\frac{\sum_{n=1}^{N} -n^{-\gamma} \log(n)}{2 \sum_{n=1}^{N} n^{-\gamma}}
\\ &=
-\frac{1}{2}\left(
\frac{\sum_{n=1}^{N} n^{-\gamma/2} \log(n)}{\sum_{n=1}^{N} n^{-\gamma/2}}
-
\frac{\sum_{n=1}^{N} n^{-\gamma} \log(n)}{\sum_{n=1}^{N} n^{-\gamma}}
\right)
\\ &=
-\frac{1}{2}
\cdot
\frac{
\left(
\sum_{n=1}^{N} n^{-\gamma/2} \log(n)
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma}
\right)
-
\left(
\sum_{n=1}^{N} n^{-\gamma/2}
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} \log(n)
\right)
}{
\left(
\sum_{n=1}^{N} n^{-\gamma/2}
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} \log(n)
\right)
}
\end{align*}$$
It is thus sufficient to show that the following proposition $\mathcal{P}(N)$ 
$$
\mathcal{P}(N) : 
\left(
\sum_{n=1}^{N} n^{-\gamma/2} \log(n)
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma}
\right)
-
\left(
\sum_{n=1}^{N} n^{-\gamma/2}
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} \log(n)
\right)
\geq 
0
\quad
\text{for all}\
\gamma \geq 0
$$
is true for all positive integers $N$. We proceed by induction on $N$. 
We see that $\mathcal{P}(1)$ is true as
$$
(n^{-\gamma/2} \log(n))(n^{-\gamma}) - (n^{-\gamma/2})(n^{-\gamma} \log(n))
=
0
$$
for all $\gamma \geq 0 $.
Now 
$$\begin{align*}
\, &
\left(
\sum_{n=1}^{N} n^{-\gamma/2} \log(n) + (N+1)^{-\gamma/2}\log(N+1)
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} + (N+1)^{-\gamma}
\right)
-
\left(
\sum_{n=1}^{N} n^{-\gamma/2} + (N+1)^{-\gamma/2} 
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} \log(n) + (N+1)^{-\gamma}\log(N+1)
\right)
\\ = &\,
\left(
\sum_{n=1}^{N} n^{-\gamma/2} \log(n) 
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} 
\right)
+
\left(
\sum_{n=1}^{N} n^{-\gamma/2} \log(n) 
\right)
(N+1)^{-\gamma}
\\ &+
(N+1)^{-\gamma/2}\log(N+1)
\left(
\sum_{n=1}^{N} n^{-\gamma} 
\right)
+
(N+1)^{-\gamma/2}\log(N+1)
(N+1)^{-\gamma}
\\
&-
\left(
\sum_{n=1}^{N} n^{-\gamma/2}  
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} \log(n)
\right)
-
\left(
\sum_{n=1}^{N} n^{-\gamma/2}  
\right)
(N+1)^{-\gamma}\log(N+1)
\\ &-
(N+1)^{-\gamma/2}
\left(
\sum_{n=1}^{N} n^{-\gamma} \log(n)
\right)
-
(N+1)^{-\gamma/2}
(N+1)^{-\gamma}\log(N+1)
\\ = &\,
\left(
\sum_{n=1}^{N} n^{-\gamma/2} \log(n) 
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} 
\right)
-
\left(
\sum_{n=1}^{N} n^{-\gamma/2}  
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} \log(n)
\right)
\\ &+
\sum_{n=1}^{N}
\log(N+1)\left(
(N+1)^{-\gamma/2}
n^{-\gamma}
-
(N+1)^{-\gamma}
n^{-\gamma/2}  
\right)
\\ &-
\sum_{n=1}^{N}
\log(n)\left(
(N+1)^{-\gamma/2}
n^{-\gamma}
-
(N+1)^{-\gamma}
n^{-\gamma/2}  
\right)
\\ = &\,
\left(
\sum_{n=1}^{N} n^{-\gamma/2} \log(n) 
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} 
\right)
-
\left(
\sum_{n=1}^{N} n^{-\gamma/2}  
\right)
\left(
\sum_{n=1}^{N} n^{-\gamma} \log(n)
\right)
\\ &+
\sum_{n=1}^{N}
\left(
  \log(N+1) - \log(n)
\right)
(N+1)^{-\gamma}n^{-\gamma}
\left(
  (N+1)^{\gamma/2} - n^{\gamma/2}
\right)
\end{align*}$$
Hence if $\mathcal{P}(N)$ is true then $\mathcal{P}(N+1)$ is true, as required.
