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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 August 7th, 2019, 01:53 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 711 Thanks: 96 Prove inequality without calculator Prove inequality: $\displaystyle \sqrt{2} +\sqrt{3} >\pi$. August 7th, 2019, 04:56 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 711 Thanks: 96 Let $\displaystyle y_x =\sqrt{2x} +\sqrt{3x} -\pi >0 \: \Rightarrow dy/dx>0.$ $\displaystyle yâ€™(1)>0 \: \Rightarrow y(1)=\sqrt{2} +\sqrt{3} -\pi >0$. August 7th, 2019, 06:19 AM   #3
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 Originally Posted by idontknow Let $\displaystyle y_x =\sqrt{2x} +\sqrt{3x} -\pi >0 \: \Rightarrow dy/dx>0.$ $\displaystyle yâ€™(1)>0 \: \Rightarrow y(1)=\sqrt{2} +\sqrt{3} -\pi >0$.
$y'(1) > 0 \implies y(1) > 0$ ... how? August 7th, 2019, 06:38 AM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 711 Thanks: 96 Here are similar examples . http://home.cc.umanitoba.ca/~farhadi/Inequalities.pdf August 7th, 2019, 07:55 AM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 3,031 Thanks: 1620 so, what if $f(x) = \sqrt{2x}+\sqrt{3x}-4$ ? $f'(1) > 0$, but $f(1) < 0$ ... ? Thanks from idontknow August 7th, 2019, 08:07 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,019 Thanks: 2252 For this problem, can it be assumed that $\pi$ = 3.141...? August 9th, 2019, 11:36 AM   #7
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 Originally Posted by idontknow Here are similar examples . http://home.cc.umanitoba.ca/~farhadi/Inequalities.pdf
Let $\displaystyle y=\sqrt{2x} +\sqrt{3x} -x\pi \:$ ,the minimum holds for $\displaystyle x=0$ , so $\displaystyle y(1)>y(0)=0$. August 9th, 2019, 06:40 PM #8 Global Moderator   Joined: Dec 2006 Posts: 21,019 Thanks: 2252 As y(1.1) = -0.156 approximately, it doesn't follow that y(1) > y(0). It's true that y(1) > y(0), but you haven't proved it. Thanks from idontknow August 11th, 2019, 06:14 AM   #9
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 Originally Posted by skipjack As y(1.1) = -0.156 approximately, it doesn't follow that y(1) > y(0). It's true that y(1) > y(0), but you haven't proved it.
Proving the inequality without calculator will be a good catch. August 14th, 2019, 02:08 AM #10 Senior Member   Joined: Dec 2015 From: somewhere Posts: 711 Thanks: 96 $\displaystyle 1>\frac{1}{x^2 +1 }$ at least for $\displaystyle 0\leq x \leq 2$. By comparing integrals : $\displaystyle \int_{0}^{y+x\pi }dx >\int_{0}^{1} (1+x^2 )^{-1} dx =\pi$. $\displaystyle y+x\pi =\sqrt{2} +\sqrt{3} >\pi .$ Tags calculator, inequality, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Elementary Math 7 March 30th, 2018 12:08 PM idontknow Elementary Math 3 February 20th, 2017 12:12 AM StillAlive Calculus 5 September 2nd, 2016 11:45 PM Albert.Teng Algebra 3 January 29th, 2013 06:48 AM pisco Trigonometry 1 June 10th, 2011 05:39 AM

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