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Prove that :

$$\frac{1}{\sqrt{x^2+yz+3}}+\frac{1}{\sqrt{y^2+zx+3}}+\frac{1}{\sqrt{z^2+xy+3}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.

I try to apply Cauchy-Buniakowski and I obtaine the followin:

$$\sum_{x,y,z}{\frac{1}{\sqrt{x^2+yz+3}}}\cdot \sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}\geq 9$$

So I have to prove that : $$\displaystyle\frac{9}{\sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.

Another trying : $$\left(\sum_{x,y,z}{\sqrt{x^2+yz+3}}\right) \leq \sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)} $$ so we have to prove that:

$$\frac{9}{\sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)}} \geq 1$$ hence:

$$3(x^2+y^2+z^2+xy+yz+zx+9) \leq 81$$ or

$$(x^2+y^2+z^2+xy+yz+zx+9) \leq 27$$ or

$$x^2+y^2+z^2+xy+yz+zx \leq 18$$

$$x^2+y^2+z^2+xy+yz+zx \leq 2\left(x^2+y^2+z^2\right) \leq 2 \cdot 9 =18.$$

Yes, it is ok :)

thanks :)

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Use Cauchy-Schwarz again: $(\sum \sqrt{x^2+yz+3})^2 \leq (\sum x^2+yz+3)(1+1+1)$ –  user27126 Jan 10 '13 at 18:30
    
I think is not ok. I try something , but don't work –  Iuli Jan 10 '13 at 18:35
    
What have you tried? –  user27126 Jan 10 '13 at 18:47
    
OK. So how did you try to proceed after what I said? Can you say in more details? –  user27126 Jan 10 '13 at 18:54
    
@Sanchez Thanks:) it is ok, it's ok your proof. –  Iuli Jan 10 '13 at 19:09

1 Answer 1

up vote 3 down vote accepted

I try to apply Cauchy-Buniakowski and I obtaine the followin:

$$\sum_{x,y,z}{\frac{1}{\sqrt{x^2+yz+3}}}\cdot \sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}\geq 9$$

So I have to prove that : $$\displaystyle\frac{9}{\sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.

$$\left(\sum_{x,y,z}{\sqrt{x^2+yz+3}}\right) \leq \sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)} $$ so we have to prove that:

$$\frac{9}{\sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)}} \geq 1$$ hence:

$$3(x^2+y^2+z^2+xy+yz+zx+9) \leq 81$$ or

$$(x^2+y^2+z^2+xy+yz+zx+9) \leq 27$$ or

$$x^2+y^2+z^2+xy+yz+zx \leq 18$$

$$x^2+y^2+z^2+xy+yz+zx \leq 2\left(x^2+y^2+z^2\right) \leq 2 \cdot 9 =18.$$

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