$1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$

Question

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.

Answer 1

Taking logs, this is $$\frac23\log(1+e^a)+\frac15\log(1+e^b)+\frac2{15}\log(1+e^c) \ge \log(1+e^{\frac23a+\frac15b+\frac2{15}c})$$ which follows from convexity of $\log(1+e^x)$ and Jensen's inequality.

Answer 2

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?