<s>[INST] 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.
 [/INST] Let $a=3/5$. We look at $\log(1+a^x)^{1/x}=\frac{\log(1+a^x)}{x}$ for positive $x$.
Differentiate, using the Quotient Rule. The denominator is safely positive. The numerator is
$$\frac{x}{1+a^x}(\log a)a^x-\log(1+a^x).$$
This is negative, since $\log a$ is. Since the logarithm is decreasing, so is the original function.
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