May 21, 2013 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,967 Prove that \(\displaystyle 5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}\).
May 22, 2013 #2 Z zzephod Well-known member Feb 3, 2013 134 anemone said: Prove that \(\displaystyle 5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}\). Click to expand... Spoiler AM-GM inequality shows that: $$\sqrt{5}+\root 3 \of{5}+\root 4 \of{5} >3\times 5^{13/36}>3\times 5^{1/3}$$ The cube of the right most term is $135>5^3$, so: $$\sqrt{5}+\root 3 \of{5}+\root 4 \of{5} >3\times 5^{1/3}>5$$ .
anemone said: Prove that \(\displaystyle 5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}\). Click to expand... Spoiler AM-GM inequality shows that: $$\sqrt{5}+\root 3 \of{5}+\root 4 \of{5} >3\times 5^{13/36}>3\times 5^{1/3}$$ The cube of the right most term is $135>5^3$, so: $$\sqrt{5}+\root 3 \of{5}+\root 4 \of{5} >3\times 5^{1/3}>5$$ .