Inequality in number theory

Question

Prove that:

(i)$5<5^{\frac{1}{2}}+5^{\frac{1}{3}}+5^{\frac{1}{4}}$

(ii)$8>8^{\frac{1}{2}}+8^{\frac{1}{3}}+8^{\frac{1}{4}}$

(iii)$n>n^{\frac{1}{2}}+n^{\frac{1}{3}}+n^{\frac{1}{4}}$ for all integer $n\geq9$

Can raising both sides to exponent $12$ help

Answer 1

(i) $5<2.2+1.7+1.4<5^{1/2}+5^{1/3}+5^{1/4}$.

(ii) $8>2.9+2+1.7>8^{1/2}+8^{1/3}+8^{1/4}$.

(iii) $n\ge3\sqrt n>n^{1/2}+n^{1/3}+n^{1/4}$ because $n \ge 9$.

Answer 2

HINT: For (ii) we have to prove that $$6>2\sqrt{2}+\sqrt[4]{8}$$ This can be written as $$6-2\sqrt{2}>\sqrt[4]{8}$$ Raise to the power $4$ we have $$3080>2112\sqrt{2}$$ Squaring we get $$565312>0$$ For (i) we get by $$AM-GM$$ $$\frac{5^{1/2}+5^{1/3}+5^{1/4}}{3}\geq \sqrt[3]{5^{13/12}}$$ so we have to Show that $$3\sqrt[3]{5^{13/12}}>5$$ this is equivalent to $$27^{12}>5^{23}$$ and this is $$150094635296999121>11920928955078125$$

Answer 3

The last one is equivalent to showing $$1 > \frac{1}{n^{\frac{1}{2}}}+\frac{1}{n^{\frac{2}{3}}}+\frac{1}{n^{\frac{3}{4}}}$$ and $$n\geq 9 \Rightarrow \frac{1}{3}\geq \frac{1}{n^{\frac{1}{2}}}$$ $$n^2\geq 9^2 > 3^3 \Rightarrow \frac{1}{3}> \frac{1}{n^{\frac{2}{3}}}$$ $$n^3\geq 9^3 > 3^4 \Rightarrow \frac{1}{3}> \frac{1}{n^{\frac{3}{4}}}$$ and the result follows.