# Hard inequality

#### algebralover

Given a,b,c>=1 and a+b+c=9.
Prove that (âˆša+âˆšb+âˆšc)^2>=ab+bc+ca.

#### 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 $$\displaystyle 9+2\sqrt{ab +bc+ac}$$.

#### 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 .

#### idontknow

$$\displaystyle a+b+c=3^{2} \: \implies \: \sqrt{a+b+c}=3\leq \sqrt{a}+\sqrt{b}+\sqrt{c}$$.

$$\displaystyle (\sqrt{a}+\sqrt{b}+\sqrt{c})^{2} \geq 9$$.
$$\displaystyle 9+2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})\geq 9$$.
$$\displaystyle \sqrt{ab}+\sqrt{ac}+\sqrt{bc}>0$$.
Is it done ?

#### tahirimanov19

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.....

1 person

#### idontknow

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 $$\displaystyle n=3$$ , $$\displaystyle x_1 =a$$ , $$\displaystyle x_2 =b$$ , $$\displaystyle x_3=c$$.

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