If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$(Sweden 1971)

Question

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.

Answer 1

Maybe the following is better.

Let $x=\sqrt{\frac{26}{3}}a$, $y=\sqrt{\frac{26}{3}}b$ and $z=\sqrt{\frac{26}{3}}c$.

Thus, $$ab+ac+bc=3$$ and $$x+y+z=\sqrt{\frac{26}{3}}(a+b+c)\geq\sqrt{\frac{26}{3}}\sqrt{3(ab+ac+bc)}=\sqrt{78}.$$ Thus, it's enough to prove that $$\sqrt{78}>\frac{35}{4}$$ or $$1248>1225$$ and we are done!

Answer 2

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.