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- Feb 14, 2012

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Prove that \(\displaystyle 5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}\).

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- Thread starter
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- #1

- Feb 14, 2012

- 3,682

Prove that \(\displaystyle 5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}\).

Prove that \(\displaystyle 5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}\).

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

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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