Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :)

share|improve this question
1  
Use Cauchy-Schwarz again: $(\sum \sqrt{x^2+yz+3})^2 \leq (\sum x^2+yz+3)(1+1+1)$ –  Sanchez 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? –  Sanchez Jan 10 '13 at 18:47
    
OK. So how did you try to proceed after what I said? Can you say in more details? –  Sanchez Jan 10 '13 at 18:54
    
@Sanchez Thanks:) it is ok, it's ok your proof. –  Iuli Jan 10 '13 at 19:09
show 3 more comments

1 Answer

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

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.