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

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

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    What did you try doing? – 2017-01-08
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    Raise each side to the power of $4$. Since both sides are positive, the direction of the inequality will be preserved. – 2017-01-08
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    @barakmanos then you have to prove that $(6 \cdot 3^\frac18-5)^4 > 3$ but why is it easier? I have it written on paper in front of me and can't see why it helps..? – 2017-01-08

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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.

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

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

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