Numbers, Roots, Cube Roots

Prove that (i) 5 < 51/2+51/3+51/45^{1/2}+5^{1/3}+5^{1/4}

(ii)11 > 111/2+111/3+111/411^{1/2}+11^{1/3}+11^{1/4}

#NumberTheory

Note by Ayush Choubey
6 years, 6 months ago

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Comments

(i)

Proof by Contradiction

Consider

5≥51/2+51/3+51/4,with lcm(2,3,4)=12, let z=51/12z12≥z6+z4+z3z9≥z3+z+1z9−z3≥z+159/12−53/12≥51/12+153/4−51/4≥51/12+151/4(5−1)≥51/12+1,square both sides5(6−25)≥51/6+2⋅51/12+165−10≥51/6+2⋅51/12+165≥51/6+2⋅51/12+11>1+2⋅1+11=14,square both sides again180>142,which is absurd \begin{aligned} 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{aligned}

Hence, 5<51/2+51/3+51/4 5 < 5^{1/2} + 5^{1/3} + 5^{1/4}

Another method is to consider the AM-GM inequality. Suppose 5≥51/2+51/3+51/4 5 \geq 5^{1/2} + 5^{1/3} + 5^{1/4}

⇒53≥51/2+51/3+51/43>51/2+1/3+1/43=513/36 \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}

⇒(53)36>513 \large \Rightarrow \left ( \frac {5}{3} \right )^{36} > 5^{13}

⇒523>336 \large \Rightarrow 5^{23} > 3^{36}

⇒524>523>336 \large \Rightarrow 5^{24} > 5^{23} > 3^{36}

⇒52>33 \large \Rightarrow 5^2 > 3^3

Which is absurd

(ii)

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

z9−z3≤z+1113/4−111/4≤111/12+1111/4(11−1)≤111/12+1,square both sides11(12−211)≤111/6+2⋅111/12+1<2+2⋅2+1=71211−22<71211<29 \begin{aligned} 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{aligned}

Square both sides yields a contradiction, thus 11<111/2+111/3+111/4 11 < 11^{1/2} + 11^{1/3} + 11^{1/4}

Alternatively, it's trivial to show that 111/2<4,111/3<3,111/4<211^{1/2} < 4, 11^{1/3} < 3, 11^{1/4} < 2

Add them up gives 111/2+111/3+111/4<911^{1/2} + 11^{1/3} + 11^{1/4} < 9

Now consider the negation for the given inequality, that is 11≤111/2+111/3+111/4<911 \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≤111/2+111/3+111/4 11 \leq 11^{1/2} + 11^{1/3} + 11^{1/4} , then

⇒113≤111/2+111/3+111/43<116+114+113312<2⋅116312 \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} }

⇒(113)12<2⋅1163 \large \Rightarrow \left (\frac {11}{3} \right )^{12} < \frac {2 \cdot 11^6}{3}

⇒(113)12<2⋅1163<3⋅1163=116 \large \Rightarrow \left (\frac {11}{3} \right )^{12} < \frac {2 \cdot 11^6}{3} < \frac {3 \cdot 11^6}{3} = 11^6

⇒116<312⇒11<32 \large \Rightarrow 11^6 < 3^{12} \Rightarrow 11 < 3^2 which is invalid.

Pi Han Goh - 6 years, 6 months ago

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NICE SOL. ....

Ayush Choubey - 6 years, 6 months ago

(2.2)2=4.84<5(2.2)^{2}=4.84<5 or5>2.2\sqrt{5}>2.2. 51/4>2.2>1.45^{1/4}>\sqrt{2.2}>1.4 since (1.4)2=1.96<2.2(1.4)^2=1.96<2.2

Hence,51/3>51/4>1.45^{1/3}>5^{1/4}>1.4. So,51/2+51/3+51/4>2.2+1.4+1.4=55^{1/2}+5^{1/3}+5^{1/4}>2.2+1.4+1.4=5.

I will prove that for any n≥9n≥9, n1/2+n1/3+n1/4<nn^{1/2}+n^{1/3}+n^{1/4}<n.

Since,n≥9n≥9,n2≥9nn^{2}≥9n or n≥3nn≥3\sqrt{n}.So,n1/2+n1/3+n1/4<3n1/2≤nn^{1/2}+n^{1/3}+n^{1/4}<3n^{1/2}≤n

Souryajit Roy - 6 years, 6 months ago

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

Mehul Arora - 5 years ago
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