Prove the following inequality $6 \cdot 3^\frac18-5 > 3^{\frac14}$

Question

So I had this homework to do in which I had to prove that:

$$6 \cdot 3^\frac18-5 > 3^{\frac14}$$ Any ideas ?

Answer 1

It's a quadratic inequality of $\sqrt[8]3$.

Id est, we have $\left(\sqrt[8]3-1\right)\left(\sqrt[8]3-5\right)<0$, which is obvious.

Answer 2

Let $x=3^{\frac18}$, observe that $1<x<5$

As $$\left(x-5\right)\left(x-1\right)<0$$ $$x^2-6x+5<0$$ $$x^2<6x-5$$

Sub $x=3^{\frac18}$.