April 11th, 2014, 09:05 AM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  show
Show that if real numbers x,y,z>0 satisfy the relation xyz+xy+yz+zx=4, then x+y+z>= 3. 
April 11th, 2014, 09:20 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,656 Thanks: 2634 Math Focus: Mainly analysis and algebra 
The first equation gives us $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3$$ Which I think turns out to be useful in this one. 
April 11th, 2014, 09:34 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
That doesn't look quite right... 
April 11th, 2014, 12:22 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,756 Thanks: 696  
April 11th, 2014, 01:27 PM  #5 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
Let $t=x+y+z$. By AM–GM, $$t^3\ =\ (x+y+z)^3\ \geqslant\ 27xyz\ \ldots\fbox1$$ By Cauchy–Schwarz, $$\begin{align*} t^2\ =\ (x+y+z)^2 &\leqslant\ (x^2+y^2+z^2)(1^2+1^2+1^2) \\\\ &=\ 3(x^2+y^2+z^2) \\\\ &=\ 3t^26(xy+yz+zx)\end{align*}$$ $\displaystyle \therefore\ t^2\ \geqslant\ 3(xy+yz+zx)\ =\ 123xyz\ \ldots\fbox2$ $\displaystyle \fbox1+9\times\fbox2$ gives $$\begin{align*} t^3+3t^2108 &\geqslant\ 0 \\\\ (t3)(t+6)^2 &\geqslant\ 0 \end{align*}$$ Since $(t+6)^2>0$, it follows that $t3\geqslant0$. QED 
April 11th, 2014, 05:32 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,656 Thanks: 2634 Math Focus: Mainly analysis and algebra 
Oops. I should answer when I have a bit of time to think more clearly!


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