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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 AMGM for $\displaystyle 9+2\sqrt{ab +bc+ac}$. 
October 1st, 2019, 02:52 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  
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: $a,b,c \ge 1. \; a+b+c=9$ $ab+bc+ca \le a^2+b^2+c^2$ $ab+bc+ca=b(a+c)+ac=b(9b)+ac$ $9b=a+c \ge 2 \sqrt{ac} \Rightarrow ac \le \dfrac{(9b)^2}{4}$ $ab+bc+ca=b(a+c)+ac \le b(9b)+\dfrac{(9b)^2}{4} = \dfrac{36b4b^2+8118b+b^2}{4}=\dfrac{81+18b3b^2}{4}$ .....Working on the next part..... 
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. 

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