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 
Math Team Joined: Jul 2011 From: Texas Posts: 3,031 Thanks: 1620  
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 $f(1) < 0$ ... ? 
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  
Senior Member Joined: Dec 2015 From: somewhere Posts: 711 Thanks: 96  Quote:
 
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.

August 11th, 2019, 06:14 AM  #9 
Senior Member Joined: Dec 2015 From: somewhere Posts: 711 Thanks: 96  
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 : $\displaystyle \int_{0}^{y+x\pi }dx >\int_{0}^{1} (1+x^2 )^{1} dx =\pi$. $\displaystyle y+x\pi =\sqrt{2} +\sqrt{3} >\pi .$ 

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