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Numbers, Roots, Cube Roots

Prove that (i) 5 < \(5^{1/2}+5^{1/3}+5^{1/4}\)

(ii)11 > \(11^{1/2}+11^{1/3}+11^{1/4}\)

Note by Ayush Choubey
2 years, 10 months ago

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(i)

Proof by Contradiction

Consider

\[ \begin{eqnarray} 5 & \geq & 5^{1/2} + 5^{1/3} + 5^{1/4}, \text{with } \text{lcm}(2,3,4) = 12 , \text{ let } z = 5^{1/12} \\ z^{12} & \geq & z^6 + z^4 + z^3 \\ z^9 & \geq & z^3 + z + 1 \\ z^9 - z^3 & \geq & z + 1 \\ 5^{9/12} - 5^{3/12} & \geq & 5^{1/12} + 1 \\ 5^{3/4} - 5^{1/4} & \geq & 5^{1/12} + 1 \\ 5^{1/4} (\sqrt{5} - 1) & \geq & 5^{1/12} + 1, \text{square both sides} \\ \sqrt{5} (6 - 2\sqrt{5}) & \geq & 5^{1/6} + 2 \cdot 5^{1/12} + 1 \\ 6 \sqrt{5} - 10 & \geq & 5^{1/6} + 2 \cdot 5^{1/12} + 1 \\ 6 \sqrt{5} & \geq & 5^{1/6} + 2 \cdot 5^{1/12} + 11 > 1 + 2 \cdot 1 + 11 = 14, \text{square both sides again}\\ 180 & > & 14^2, \text{which is absurd} \\ \end{eqnarray} \]

Hence, \( 5 < 5^{1/2} + 5^{1/3} + 5^{1/4} \)

Another method is to consider the AM-GM inequality. Suppose \( 5 \geq 5^{1/2} + 5^{1/3} + 5^{1/4} \)

\( \LARGE \Rightarrow \frac {5}{3} \geq \frac {5^{1/2} + 5^{1/3} + 5^{1/4} }{3} > \sqrt[3]{ 5^{1/2 + 1/3+1/4} } = 5^{13/36} \)

\( \large \Rightarrow \left ( \frac {5}{3} \right )^{36} > 5^{13} \)

\( \large \Rightarrow 5^{23} > 3^{36} \)

\( \large \Rightarrow 5^{24} > 5^{23} > 3^{36} \)

\( \large \Rightarrow 5^2 > 3^3 \)

Which is absurd


(ii)

Likewise, using the same method above (first), we have

\[ \begin{eqnarray} z^9 - z^3 & \leq & z + 1 \\ 11^{3/4} - 11^{1/4} & \leq & 11^{1/12} + 1 \\ 11^{1/4} ( \sqrt{11} - 1) & \leq & 11^{1/12} + 1, \text{square both sides} \\ \sqrt{11} (12 - 2\sqrt{11}) & \leq & 11^{1/6} + 2 \cdot 11^{1/12} + 1 < 2 + 2 \cdot 2 + 1 = 7 \\ 12 \sqrt{11} - 22 < 7 \\ 12\sqrt{11} < 29 \\ \end{eqnarray} \]

Square both sides yields a contradiction, thus \( 11 < 11^{1/2} + 11^{1/3} + 11^{1/4} \)

Alternatively, it's trivial to show that \(11^{1/2} < 4, 11^{1/3} < 3, 11^{1/4} < 2 \)

Add them up gives \(11^{1/2} + 11^{1/3} + 11^{1/4} < 9 \)

Now consider the negation for the given inequality, that is \(11 \leq 11^{1/2} + 11^{1/3} + 11^{1/4} < 9 \) which is absurd.

Another method is to apply the Power Mean Inequality.

Suppose that \( 11 \leq 11^{1/2} + 11^{1/3} + 11^{1/4} \), then

\( \LARGE \Rightarrow \frac {11}{3} \leq \frac {11^{1/2} + 11^{1/3} + 11^{1/4}}{3} < \sqrt[12] { \frac { 11^6 + 11^4 + 11^3}{3} } < \sqrt[12] { \frac {2 \cdot 11^6}{3} } \)

\( \large \Rightarrow \left (\frac {11}{3} \right )^{12} < \frac {2 \cdot 11^6}{3} \)

\( \large \Rightarrow \left (\frac {11}{3} \right )^{12} < \frac {2 \cdot 11^6}{3} < \frac {3 \cdot 11^6}{3} = 11^6 \)

\( \large \Rightarrow 11^6 < 3^{12} \Rightarrow 11 < 3^2 \) which is invalid.

Pi Han Goh - 2 years, 10 months ago

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NICE SOL. ....

Ayush Choubey - 2 years, 10 months ago

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\((2.2)^{2}=4.84<5\) or\(\sqrt{5}>2.2\). \(5^{1/4}>\sqrt{2.2}>1.4\) since \((1.4)^2=1.96<2.2\)

Hence,\(5^{1/3}>5^{1/4}>1.4\). So,\(5^{1/2}+5^{1/3}+5^{1/4}>2.2+1.4+1.4=5\).

I will prove that for any \(n≥9\), \(n^{1/2}+n^{1/3}+n^{1/4}<n\).

Since,\(n≥9\),\(n^{2}≥9n\) or \(n≥3\sqrt{n}\).So,\(n^{1/2}+n^{1/3}+n^{1/4}<3n^{1/2}≤n\)

Souryajit Roy - 2 years, 10 months ago

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This is the perfect solution. This was asked in INMO, as far as I remember.

Mehul Arora - 1 year, 4 months ago

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