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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)

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.

- 2 years, 10 months ago

NICE SOL. ....

- 2 years, 10 months ago

$$(2.2)^{2}=4.84<5$$ 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,$$n^{1/2}+n^{1/3}+n^{1/4}<3n^{1/2}≤n$$

- 2 years, 10 months ago

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

- 1 year, 4 months ago