Show sequence ${a_n} = \sqrt[n]{{{3^n} + {5^n}}}$ is monotone decreasing (a)  Show that sequence ${a_n} = \sqrt[n]{{{3^n} + {5^n}}}$ is monotone decreasing 
Proof
Let ${a_n} = \sqrt[n]{{{3^n} + {5^n}}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^n} + 1} \right]^{\frac{1}{n}}}$  then 
${a_k} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{k}}}$
${a_{k + 1}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$ 
We Know   ${\left( {\frac{3}{5}} \right)^k} \geqslant {\left( {\frac{3}{5}} \right)^{k + 1}}$ 
${\left( {\frac{3}{5}} \right)^k} + 1 \geqslant {\left( {\frac{3}{5}} \right)^{k + 1}} + 1$ 
${\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{{k + 1}}}} \geqslant {\left[ {{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$ 
${\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{k}}} \geqslant {\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{{k + 1}}}} \geqslant $ 
${\left[ {{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$
$5{\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{k}}} \geqslant 5{\left
[{{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$ 
${a_k} \geqslant {a_{k + 1}}$ 
So ${a_n} = \sqrt[n]{{{3^n} + {5^n}}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^n} + 1} \right]^{\frac{1}{n}}}$ is monotone decreasing .
I not sure it True or False ,I hope someone help me thank.

Hint: Taking the log-derivative of $5\left(1+(3/5)^n\right)^{1/n}$ yields
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}n}\left(\log(5)+\frac1n\log\left(1+(3/5)^n\right)\right)
&=\frac1n\frac{\color{#C00000}{\log(3/5)}(3/5)^n}{1+(3/5)^n}\color{#C00000}{-\frac1{n^2}}\log\left(1+(3/5)^n\right)\end{align}
$$

I've just noticed that André's answer uses the quotient rule while mine uses product rule; otherwise, mine is not much different. To give added value to my answer, I feel I should give an answer that does not use calculus.
Bernoulli's Inequality, the rational version of which is proven here, shows
$$
\begin{align}
\left[1+\left(\frac35\right)^n\right]^{\Large\frac{n+1}{n}}
&\ge1+\frac{n+1}{n}\left(\frac35\right)^n\\
&\ge1+\frac35\left(\frac35\right)^n\\
&=1+\left(\frac35\right)^{n+1}\\
\left[1+\left(\frac35\right)^n\right]^{\Large\frac1n}
&\ge\left[1+\left(\frac35\right)^{n+1}\right]^{\Large\frac1{n+1}}\\
\left[5^n+3^n\vphantom{3^{n+1}}\right]^{\Large\frac1n}
&\ge\left[5^{n+1}+3^{n+1}\right]^{\Large\frac1{n+1}}
\end{align}
$$
