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August 7th, 2019, 01:53 AM   #1
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Prove inequality without calculator

Prove inequality: $\displaystyle \sqrt{2} +\sqrt{3} >\pi $.
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August 7th, 2019, 04:56 AM   #2
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Let $\displaystyle y_x =\sqrt{2x} +\sqrt{3x} -\pi >0 \: \Rightarrow dy/dx>0.$

$\displaystyle y’(1)>0 \: \Rightarrow y(1)=\sqrt{2} +\sqrt{3} -\pi >0 $.
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August 7th, 2019, 06:19 AM   #3
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Quote:
Originally Posted by idontknow View Post
Let $\displaystyle y_x =\sqrt{2x} +\sqrt{3x} -\pi >0 \: \Rightarrow dy/dx>0.$

$\displaystyle y’(1)>0 \: \Rightarrow y(1)=\sqrt{2} +\sqrt{3} -\pi >0 $.
$y'(1) > 0 \implies y(1) > 0$ ... how?
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August 7th, 2019, 06:38 AM   #4
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Here are similar examples .
http://home.cc.umanitoba.ca/~farhadi/Inequalities.pdf
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August 7th, 2019, 07:55 AM   #5
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so, what if $f(x) = \sqrt{2x}+\sqrt{3x}-4$ ?

$f'(1) > 0$, but $f(1) < 0$

... ?
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August 7th, 2019, 08:07 AM   #6
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For this problem, can it be assumed that $\pi$ = 3.141...?
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August 9th, 2019, 11:36 AM   #7
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Let $\displaystyle y=\sqrt{2x} +\sqrt{3x} -x\pi \: $ ,the minimum holds for $\displaystyle x=0$ , so $\displaystyle y(1)>y(0)=0 $.
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August 9th, 2019, 06:40 PM   #8
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As y(1.1) = -0.156 approximately, it doesn't follow that y(1) > y(0). It's true that y(1) > y(0), but you haven't proved it.
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August 11th, 2019, 06:14 AM   #9
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As y(1.1) = -0.156 approximately, it doesn't follow that y(1) > y(0). It's true that y(1) > y(0), but you haven't proved it.
Proving the inequality without calculator will be a good catch.
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August 14th, 2019, 02:08 AM   #10
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$\displaystyle 1>\frac{1}{x^2 +1 } $ at least for $\displaystyle 0\leq x \leq 2$.
By comparing integrals :
$\displaystyle \int_{0}^{y+x\pi }dx >\int_{0}^{1} (1+x^2 )^{-1} dx =\pi$.
$\displaystyle y+x\pi =\sqrt{2} +\sqrt{3} >\pi .$
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