Determine whether there exists $c\in\mathbb{R}$ such that $x^3 > 2^{x/2}$ for $x \in [c, \infty)$.
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2You do not "prove if something". You probably mean "Determine whether something." Said determination may naturally include a proof. – Andrés E. Caicedo Jul 17 '16 at 14:05
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Well the inequality holds for $x=2$, but not for $x=1$. It does not hold for $x\geq 29.22$. – smcc Jul 17 '16 at 14:13
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Changed the question from 'Prove that' to 'Determine whether'. – Himanshu Ahuja Jul 17 '16 at 14:19
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@HimanshuAhuja: there were some mistakes in the way you had written the question. I edited. Please read over the edits to understand the errors. – parsiad Jul 17 '16 at 14:20
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@par : Sorry for the errors. Thank you for the edit. – Himanshu Ahuja Jul 17 '16 at 14:24
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I suppose you could take logs of both sides, giving the question $3 \ln x \gt x \ln \sqrt{2}$? This looks like you need to determine $x$ st $\ln x > x \ln 2^{1/6}$ or $\ln x \gt 0.116 x$. – jim Jul 17 '16 at 14:34
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In such cases, you should try plotting these graphs or trying various values of $x$ first. You'll find your answer pretty easily.
Daishisan
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Hint: there exists a $c$ such that for all $x \geq c$, $x^3 < 2^{x/2}$ (i.e. the function on the right grows faster than the function on the left). See also this.
Hint 2: this question.
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Can it be solved without plotting a graph? Through some other inequality or identity? – Himanshu Ahuja Jul 17 '16 at 14:17
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Yes, I was just providing you with a hint on how to proceed. Try to find a $c$ as described in my hint. – parsiad Jul 17 '16 at 14:18
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