<s>[INST] If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$(Sweden 1971) If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$
I was trying to do the question above using homogenization. I was trying to do it as follows:
$\sqrt{xy+yz+xz}=\sqrt{26}$
Hence $x+y+z>\frac{35}{4*\sqrt{26}}*\sqrt{xy+yz+xz}$
So it is enough to prove that $16*26(x^2+y^2+z^2+2xy+2yz+2xz)>35*35*(xy+yz+xz)$. Which is true since $x^2+y^2+z^2+2xy+2yz+2xz\ge 3(xy+yz+xz)$ Could you please share other simple ways to solve the question, as I am intrigued by its simplicity.
 [/INST] We can use AM-GM inequality to get
$$26 = xy + yz + xz \leq \frac{x^2+y^2}{2} + \frac{y^2+z^2}{2} + \frac{x^2+z^2}{2} = x^2+y^2+z^2$$
Then add the two inequalities
$$x^2+y^2+z^2+2xy+2yz+2xz = (x+y+z)^2 > 78$$
$$\implies x+y+z > \sqrt{78} > \frac{35}{4}$$
Oddly enough, this only relies on $x+y+z$ being positive, and we never used the more restrictive range.
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