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September 28th, 2019, 05:31 PM   #1
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Smile Hard inequality

Given a,b,c>=1 and a+b+c=9.
Prove that (√a+√b+√c)^2>=ab+bc+ca.
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September 30th, 2019, 02:23 AM   #2
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$\displaystyle a+b+c +2(\sqrt{ab} +\sqrt{ac}+\sqrt{bc} )=9+2(\sqrt{ab} +\sqrt{ac}+\sqrt{bc} )\geq 9+2\sqrt{ab +bc+ac}$.
Now apply AM-GM for $\displaystyle 9+2\sqrt{ab +bc+ac}$.
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October 1st, 2019, 02:52 AM   #3
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Quote:
Originally Posted by idontknow View Post
$\displaystyle a+b+c +2(\sqrt{ab} +\sqrt{ac}+\sqrt{bc} )=9+2(\sqrt{ab} +\sqrt{ac}+\sqrt{bc} )\geq 9+2\sqrt{ab +bc+ac}$.
Now apply AM-GM for $\displaystyle 9+2\sqrt{ab +bc+ac}$.
My mistake , it proves nothing , maybe someone else can finish it .
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October 1st, 2019, 04:08 AM   #4
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$\displaystyle a+b+c=3^{2} \: \implies \: \sqrt{a+b+c}=3\leq \sqrt{a}+\sqrt{b}+\sqrt{c}$.

$\displaystyle (\sqrt{a}+\sqrt{b}+\sqrt{c})^{2} \geq 9$.
$\displaystyle 9+2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})\geq 9$.
$\displaystyle \sqrt{ab}+\sqrt{ac}+\sqrt{bc}>0$.
Is it done ?
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October 2nd, 2019, 01:22 AM   #5
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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
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October 3rd, 2019, 04:33 AM   #6
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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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