Friday, 26 August 2016

algebra precalculus - Solution of equation fracxcdot2014frac1x+frac1xcdot2014x2=2014




Solution of equation x20141x+1x2014x2=2014




MyTry:: Clearly Here x>0, Now Using A.MG.M




So Here x20141x>0 and 1x2014x>0 .



So (x20141x+1x2014x2)(x20141x1x2014x)12=2014(x+1x)=2014



And equality Hold , When x=1xx=1



Can we solve it without Using A.MG.M



If Yes, Then please explain me.




Thanks


Answer



Applying AM-GM probably gives the most elegant solution (and indeed the presentation is quite suggestive of AM-GM), but this can also be solved by using an idea behind AM-GM. Noting first that x0, we have:



x20141x+1x2014x2=2014x20141x1+1x2014x1=2(x20141x11x2014x1)2+220141x+x2=2(x20141x11x2014x1)2+22014(1xx)2=222014(1xx)222014(1xx)21(1xx)20(1xx)2=01xx=01x=xx=1



where we have used the fact that x0 several times.



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