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 September 28th, 2019, 05:31 PM #1 Newbie   Joined: Sep 2019 From: In a far far away land Posts: 1 Thanks: 0 Hard inequality Given a,b,c>=1 and a+b+c=9. Prove that (√a+√b+√c)^2>=ab+bc+ca. September 30th, 2019, 02:23 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 $\displaystyle a+b+c +2(\sqrt{ab} +\sqrt{ac}+\sqrt{bc} )=9+2(\sqrt{ab} +\sqrt{ac}+\sqrt{bc} )\geq 9+2\sqrt{ab +bc+ac}$. Now apply AM-GM for $\displaystyle 9+2\sqrt{ab +bc+ac}$. October 1st, 2019, 02:52 AM   #3
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 Originally Posted by idontknow $\displaystyle a+b+c +2(\sqrt{ab} +\sqrt{ac}+\sqrt{bc} )=9+2(\sqrt{ab} +\sqrt{ac}+\sqrt{bc} )\geq 9+2\sqrt{ab +bc+ac}$. Now apply AM-GM for $\displaystyle 9+2\sqrt{ab +bc+ac}$.
My mistake , it proves nothing , maybe someone else can finish it . October 1st, 2019, 04:08 AM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 $\displaystyle a+b+c=3^{2} \: \implies \: \sqrt{a+b+c}=3\leq \sqrt{a}+\sqrt{b}+\sqrt{c}$. $\displaystyle (\sqrt{a}+\sqrt{b}+\sqrt{c})^{2} \geq 9$. $\displaystyle 9+2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})\geq 9$. $\displaystyle \sqrt{ab}+\sqrt{ac}+\sqrt{bc}>0$. Is it done ? October 2nd, 2019, 01:22 AM #5 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle What we know so far: $a,b,c \ge 1. \; a+b+c=9$ $ab+bc+ca \le a^2+b^2+c^2$ $ab+bc+ca=b(a+c)+ac=b(9-b)+ac$ $9-b=a+c \ge 2 \sqrt{ac} \Rightarrow ac \le \dfrac{(9-b)^2}{4}$ $ab+bc+ca=b(a+c)+ac \le b(9-b)+\dfrac{(9-b)^2}{4} = \dfrac{36b-4b^2+81-18b+b^2}{4}=\dfrac{81+18b-3b^2}{4}$ .....Working on the next part..... Thanks from idontknow October 3rd, 2019, 04:33 AM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 In simple words the hint is to apply $\displaystyle \sqrt{x_1 +...+x_n}<\sqrt{x_1 }+...+\sqrt{x_n } \;$ ; $\displaystyle x_1 , x_2 , ... x_n >0$. Set $\displaystyle n=3$ , $\displaystyle x_1 =a$ , $\displaystyle x_2 =b$ , $\displaystyle x_3=c$. Last edited by idontknow; October 3rd, 2019 at 04:35 AM. Tags hard, help me, inequality Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post matisolla Algebra 5 June 4th, 2015 05:59 PM yo79 Math Events 1 January 15th, 2013 01:10 AM henry12345 Trigonometry 2 July 7th, 2012 02:44 AM ime Algebra 0 August 23rd, 2009 02:37 AM caugust Applied Math 0 February 23rd, 2007 06:39 PM

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