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August 14th, 2019, 10:13 AM   #11
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Quote:
 Originally Posted by idontknow $\displaystyle 1>\frac{1}{x^2 +1 }$ at least for $\displaystyle 0\leq x \leq 2$.
That's incorrect. That inequality holds for all real values of $x$ except 0.

Quote:
 Originally Posted by idontknow $\displaystyle \int_{0}^{1} (1+x^2 )^{-1} dx =\pi$.
That's incorrect. That integral's value is $\pi/4$.

 August 15th, 2019, 12:34 AM #12 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 My mistake . I will seek for solution.
 August 16th, 2019, 02:54 AM #13 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 $\displaystyle x>\frac{2}{1+x^2 } \:$, $\displaystyle x>0$. 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, 03:23 AM   #14
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Quote:
 Originally Posted by idontknow $\displaystyle x>\frac{2}{1+x^2}\:$, $\displaystyle x>0$.
For $x < 1$, $\displaystyle x < \frac{2}{1 + x^2}$.

 August 17th, 2019, 07:49 AM #15 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 Somehow the proof comes in short-terms : 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 $\displaystyle sgn(x)=-1+\lim_{k\rightarrow \infty }\frac{4x-4\pi e^{2k(x-\pi)}}{2\pi e^{2k\pi }+2(x-\pi )e^{2k\pi }}=-1+\lim_{k\rightarrow \infty } \frac{4(x-\pi)}{2\pi e^{2k\pi -2sqrt(3) -2sqrt(2) }+2(sqrt(2)+sqrt(3)}=-1+\frac{2sqrt(3)+2 sqrt(2) }{sqrt(3)+sqrt(2) }=-1+ \frac{2(x-\pi)}{x-\pi }=2-1=1>0$. Now $\displaystyle sgn(x)>0 \: \equiv \: x>0$ (both are positive sign) proves the inequality. Last edited by idontknow; August 17th, 2019 at 07:51 AM.
 August 25th, 2019, 04:47 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 Thanks from idontknow
 August 25th, 2019, 08:44 PM #17 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 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 : $\displaystyle b>a \equiv sgn(b-a)>0$ and $\displaystyle b0 and see if sgn(x) is positive , if not then x<0 . While evaluating H(x) the Lâ€™hopitalâ€™s rule is a shortcut.$\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 08:46 PM.

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