For $a,b>0$, $c>1$, and $n=3,4,\dots$ is $|ca^n+(-b)^n|\leq(ca^2+b^2)^{n/2}$ always true? Let $a,b>0$, $c>1$, and $n=3,4,\dots$, then is
\begin{equation}
|ca^n+(-b)^n|\leq(ca^2+b^2)^{n/2},
\end{equation}
true for all possible combinations of $a$, $b$, $c$, and $n$?
 A: Hint: prove first that $\,\big(c+x^n\big)^2 \le \big(c+x^2\big)^n\,$ for $\,c \ge 1, x \ge 0, n \ge 2\,$, using for example the binomial expansion, noting that $\,\color{red}{c^2} \le \color{red}{c^n}\,$ and $\,\color{blue}{2cx^n}\,$ is no larger than either of the two blue terms on the RHS depending on whether $\,x \ge 1\,$ (when $\,x^n \le x^{2n-2}\,$) or $\,x \le 1\,$ (when $\,x^n \le x^{2}\,$):
$$\require{cancel}
\color{red}{c^2} + \color{blue}{2cx^n} + \cancel{x^{2n}} \le \color{red}{c^n} + \color{blue}{n \cdot c^{n-1}x^2}  + \ldots + \color{blue}{n \cdot c x^{2n-2}} + \cancel{x^{2n}}
$$
Then, use the above for $\,x=b/a\,$, multiply the inequality by $\,a^{2n}\,$ etc.
A: This is true and a pretty weak statement. Here is a much simplified and direct proof for the setup and equivalent general proposition of Alex Francisco

Given $x_1, \cdots, x_m \geqslant 0,\,λ_1, \cdots, λ_m \geqslant 1$ and $t>1$, then
  $$\sum_{k=1}^m \lambda_kx_k^t<\bigg(\sum_{k=1}^m \lambda_kx_k\bigg)^t.$$

Proof: $$\frac{\sum_{k=1}^m \lambda_kx_k^t}{\bigg(\sum_{k=1}^m \lambda_kx_k\bigg)^t}=\sum_{k=1}^m \lambda_k^{1-t}\alpha_k^t<\sum_{k=1}^m \alpha_k=1,$$
since obviously $\lambda_k^{1-t}<1$, and $\alpha_k:=\frac{\lambda_kx_k}{\sum_{k=1}^m \lambda_kx_k}\in[0,1]\implies \alpha_k^t<\alpha_k \bigwedge \sum_{k=1}^m \alpha_k=1$. 
A: Yes, it is true. Since $|ca^n + (-b)^n| \leqslant |ca^n| + |(-b)^n| = ca^n + b^n$, it suffices to prove that$$
ca^n + b^n \leqslant (ca^2 + b^2)^{\frac{n}{2}},
$$
i.e.$$
(ca^n + b^n)^{\frac{1}{n}} \leqslant (ca^2 + b^2)^{\frac{1}{2}}.
$$
Next, a more general proposition will be proved. For simplicity, take logarithms first.

Given $x_1, \cdots, x_m \geqslant 0$ and $λ_1, \cdots, λ_m \geqslant 1$, then$$
f(t) = \frac{1}{t} \ln\left(\sum_{k = 1}^m λ_k x_k^t\right)
$$
  is decreasing for $t > 0$.

Proof: Without loss of generality, assume $x_1, \cdots, x_m > 0$. Since$$
f'(t) = -\frac{1}{t^2} \ln\left(\sum_{k = 1}^m λ_k x_k^t\right) + \frac{1}{t} \frac{\sum\limits_{k = 1}^m λ_k x_k^t \ln x_k}{\sum\limits_{k = 1}^m λ_k x_k^t},
$$
then for $t > 0$,$$
f'(t) \leqslant 0 \Longleftrightarrow \left(\sum_{k = 1}^m λ_k x_k^t\right) \ln\left(\sum_{k = 1}^m λ_k x_k^t\right) \geqslant t \sum_{k = 1}^m λ_k x_k^t \ln x_k.
$$
Note that $λ_1, \cdots, λ_m \geqslant 1$, thus\begin{align*}
&\mathrel{\phantom{=}} \left(\sum_{k = 1}^m λ_k x_k^t\right) \ln\left(\sum_{k = 1}^m λ_k x_k^t\right) = \sum_{k = 1}^m λ_k x_k^t \ln\left(\sum_{j = 1}^m λ_j x_j^t\right)\\
&\geqslant \sum_{k = 1}^m λ_k x_k^t \ln(λ_k x_k^t) \geqslant \sum_{k = 1}^m λ_k x_k^t \ln(x_k^t) = t \sum_{k = 1}^m λ_k x_k^t \ln x_k.
\end{align*}
Therefore $f(t)$ is decreasing for $t > 0$.
Now, set $m = 2$ and $x_1 = a$, $x_2 = b$, $λ_1 = c$, $λ_2 = 1$, by $f(n) \leqslant f(2)$ there is$$
\frac{1}{n} \ln(ca^n + b^n) \leqslant \frac{1}{2} \ln(ca^2 + b^2),
$$
i.e.$$
(ca^n + b^n)^{\frac{1}{n}} \leqslant (ca^2 + b^2)^{\frac{1}{2}}.
$$
