Is 5 Less Than the Sum of Square, Cube, and Fourth Roots of 5?

[anemone]

Prove that $$5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}$$.

 

[zzephod]

Prove that $$5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}$$.

Spoiler

AM-GM inequality shows that:

$$\sqrt{5}+\root 3 \of{5}+\root 4 \of{5} >3\times 5^{13/36}>3\times 5^{1/3}$$

The cube of the right most term is $135>5^3$, so:

$$\sqrt{5}+\root 3 \of{5}+\root 4 \of{5} >3\times 5^{1/3}>5$$

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