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 October 22nd, 2016, 02:32 AM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 exponential inequality values of parameter $a$ for which the inequality $a\cdot9^x+4(a-1)3^x+a>1$ is true
 October 22nd, 2016, 05:45 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 This is sort of a trick question. $9^x > 0 < 4 * 3^x\ for\ all\ real\ x.$ Can you take it from here? Thanks from panky
 October 22nd, 2016, 06:10 AM #3 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 To jeffm would you like to explain me in detail, Thanks
 November 8th, 2016, 03:22 AM #4 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 Thanks Jeffm1 got it. Denoting $3^x=t>0$, we need $a>\dfrac {4t+1}{t^2 + 4t+1}$ for all $t>0$. Since the function $t\mapsto \dfrac {4t+1}{t^2 + 4t+1}$ is decreasing on $[0,\infty)$, we need $a\geq \lim_{t\to 0} \dfrac {4t+1}{t^2 + 4t+1} = 1$, and so indeed the answer is $a\in [1,\infty)$

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