August 14th, 2019, 11:13 AM  #11  
Global Moderator Joined: Dec 2006 Posts: 21,127 Thanks: 2336  Quote:
That's incorrect. That integral's value is $\pi/4$.  
August 15th, 2019, 01:34 AM  #12 
Senior Member Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math 
My mistake . I will seek for solution.

August 16th, 2019, 03:54 AM  #13 
Senior Member Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math 
$\displaystyle x>\frac{2}{1+x^2 } \: $, $\displaystyle x>0$. Let $\displaystyle c=\sqrt{2^{1/2} + 3^{1/2} }>1$. $\displaystyle \int_{0}^{c} xdx > \int_{0}^{1} \frac{2dx}{1+x^2 }=\pi /2 $. $\displaystyle \frac{\sqrt{2} + \sqrt{3}}{2}>\pi /2$. 
August 16th, 2019, 04:23 AM  #14 
Global Moderator Joined: Dec 2006 Posts: 21,127 Thanks: 2336  
August 17th, 2019, 08:49 AM  #15 
Senior Member Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math 
Somehow the proof comes in shortterms : Let $\displaystyle x=\sqrt{3} +\sqrt{2} \pi >0 \: \equiv \: sgn(x)>0$. $\displaystyle sgn(x)=1+2H(x)=1+\lim_{k\rightarrow \infty } \frac{2}{1+e^{kx }}.$ H(x) is the heaviside function.2H(x)1=sgn(x). https://en.m.wikipedia.org/wiki/Sign_function $\displaystyle sgn(x)=1+\lim_{k\rightarrow \infty }\frac{4x4\pi e^{2k(x\pi)}}{2\pi e^{2k\pi }+2(x\pi )e^{2k\pi }}=1+\lim_{k\rightarrow \infty } \frac{4(x\pi)}{2\pi e^{2k\pi 2sqrt(3) 2sqrt(2) }+2(sqrt(2)+sqrt(3)}=1+\frac{2sqrt(3)+2 sqrt(2) }{sqrt(3)+sqrt(2) }=1+ \frac{2(x\pi)}{x\pi }=21=1>0$. Now $\displaystyle sgn(x)>0 \: \equiv \: x>0$ (both are positive sign) proves the inequality. Last edited by idontknow; August 17th, 2019 at 08:51 AM. 
August 25th, 2019, 05:47 PM  #16 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 393 Thanks: 71 
This problem can be solved with arithmetic only. First calculate âˆš2 and âˆš3 to three decimal places. Do this by using the square root algorithm. This used to be taught to grade 8 students. You can find out from the link below. Skip by the trial and error method and begin reading from the title "Finding square roots using an algorithm." https://www.homeschoolmath.net/teach...algorithm.php So to solve the problem 1. Calculate âˆš2 to three decimal places. That would be 1.414 (which, of course, is less than the actual square root). 2. Calculate âˆš3 to three decimal places. That is 1.732 (also less than the actual square root) 3. Add these two values. They add to 3.146 The value of pi to four decimal places is 3.1416 (this is greater than the actual value). 4. The fact that 3.146>3.1416 implies that âˆš2+âˆš3>pi 
August 25th, 2019, 09:44 PM  #17 
Senior Member Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math 
I know but I found it useful to apply the sgn function involving contradictions. For example : Compare a and b using sgn function by following statements : $\displaystyle b>a \equiv sgn(ba)>0$ and $\displaystyle b<a \equiv sgn(ba)<0$. Where sgn(x)=1+2H(x) . Also byb contradictions : suppose x>0 and see if sgn(x) is positive , if not then x<0 . While evaluating H(x) the Lâ€™hopitalâ€™s rule is a shortcut. $\displaystyle H(x)=\lim_{s\rightarrow \infty } e^{sx} (1+e^{sx} )^{1}.$ Test this method for $\displaystyle 22/7 > \pi $ (without integrals) Last edited by idontknow; August 25th, 2019 at 09:46 PM. 

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