<s>[INST] $1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$ The inequality $1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$ is true for all $a,b,c\in\mathbb{R}$?
I've tried to use the Bernoulli inequality $(1+k)^n \geq 1+kn$ but my inequality don't satisfy the conditions. 
My other apporach was by the binomial theorem $(1+x)^n = \sum_{k=0}^n {n \choose k}x^k$ but with non integer power I don't see a way here to solve it.
 [/INST] Hint: use Holder inequality we have
$$(1+e^a)^{10}\cdot (1+e^b)^3\cdot(1+e^c)^{2}\ge (1+\sqrt[15]{e^{10a}\cdot e^{3b}\cdot e^{2c}})^{15}$$
I add $n=3$ Holder inequality proof,ie:
$$(a^3+b^3+c^3)(p^3+q^3+r^3)(u^3+v^3+w^3)\ge (apu+bqv+crw)^3,a,b,c,p,q,r,u,v,w>0$$
since
$$\sum_{cyc}\left(\dfrac{a^3}{a^3+b^3+c^3}+\dfrac{p^3}{p^3+q^3+r^3}+\dfrac{u^3}{u^3+v^3+w^3}\right)=3$$
use AM-GM inequality we have
$$3\ge 3\sum_{cyc}\dfrac{apu}{\sqrt[3]{(a^3+b^3+c^3)(u^3+v^3+w^3)(p^3+q^3+r^3)}}$$
so
$$(a^3+b^3+c^3)(p^3+q^3+r^3)(u^3+v^3+w^3)\ge (apu+bqv+crw)^3$$
then you understand How to prove    general Holder inequality?
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