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 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: 7,681 Thanks: 2659 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... Thanks from v8archie
April 11th, 2014, 12:22 PM   #4
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 Originally Posted by v8archie 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.
Right side is wrong. Should be 4/(xyz) - 1

 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^2-6(xy+yz+zx)\end{align*} $\displaystyle \therefore\ t^2\ \geqslant\ 3(xy+yz+zx)\ =\ 12-3xyz\ \ldots\fbox2$ $\displaystyle \fbox1+9\times\fbox2$ gives \begin{align*} t^3+3t^2-108 &\geqslant\ 0 \\\\ (t-3)(t+6)^2 &\geqslant\ 0 \end{align*} Since $(t+6)^2>0$, it follows that $t-3\geqslant0$. QED Thanks from v8archie
 April 11th, 2014, 05:32 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 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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