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$$\displaystyle 4^{\tfrac12} \ + \ 4^{\tfrac13} \ + \ 4^{\tfrac14} \ \ \ versus \ \ \ 5$$ or, in other symbols, $$\displaystyle \sqrt{4} \ + \ \sqrt[3]{4} \ + \ \sqrt[4]{4} \ \ \ versus \ \ \ 5$$ Without using a calculator or computer, please show the steps to determine it. Please hide your solution(s).

greg1313 said: MacstersUndead, what is your conclusion about the relative sizes of the two sides? Click to expand... Spoiler $$\displaystyle 4^{1/2} + 4^{1/3} + 4^{1/4} > 5$$, because 1681
MacstersUndead said: Spoiler $$\displaystyle 4^{1/2} + 4^{1/3} + 4^{1/4} > 5 \iff \\ 2 + 4^{1/3} + 4^{1/4} > 5 \iff \\ 4^{1/3} + 4^{1/4} > 3 \iff \\ 4^{1/3} > 3 - 2^{1/2} \iff \\ 4 > (3-2^{1/2})^3 = 45 - 29\sqrt{2} \iff \\ -41 > -29\sqrt{2} \iff \\ 41 by 0 like 0 dislike MacstersUndead said: Spoiler \(\displaystyle 4^{1/2} + 4^{1/3} + 4^{1/4} > 5 \iff \\ 2 + 4^{1/3} + 4^{1/4} > 5 \iff \\ 4^{1/3} + 4^{1/4} > 3 \iff \\ 4^{1/3} > 3 - 2^{1/2} \iff \\ 4 > (3-2^{1/2})^3 = 45 - 29\sqrt{2} \iff \\ -41 > -29\sqrt{2} \iff \\ 41 by 0 like 0 dislike Spoiler \(\displaystyle 4^{1/2} + 4^{1/3} + 4^{1/4} > 5 \iff \\ 2 + 4^{1/3} + 4^{1/4} > 5 \iff \\ 4^{1/3} + 4^{1/4} > 3 \iff \\ 4^{1/3} > 3 - 2^{1/2} \iff \\ 4 > (3-2^{1/2})^3 = 45 - 29\sqrt{2} \iff \\ -41 > -29\sqrt{2} \iff \\ 41 by 0 like 0 dislike Jomo said: Spoiler \(\displaystyle \sqrt 4 = 2\ and\ \sqrt[4]{4} = \sqrt 2\approx 1.414$$ The sum so far is ~3.414 We have ~1.6 before going over 5. The question is whether or not $$\displaystyle \sqrt[3]{4} by 0 like 0 dislike Jomo said: Spoiler \(\displaystyle \sqrt 4 = 2\ and\ \sqrt[4]{4} = \sqrt 2\approx 1.414$$ The sum so far is ~3.1414 We have ~1.8 before going over 5. The question is whether or not $$\displaystyle \sqrt[3]{4} by 0 like 0 dislike Spoiler \(\displaystyle \sqrt 4 = 2\ and\ \sqrt[4]{4} = \sqrt 2\approx 1.414$$ The sum so far is ~3.1414 We have ~1.8 before going over 5. The question is whether or not \(\displaystyle \sqrt[3]{4}