# Problem 6. (3 points)

Positive numbers $x, y, z$ are such that $x y + y z + x z = 12$.

Find the smallest possible value of $x + y + z$.

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# Answer: 6

## Solution:

By adding the inequalities $x^{2}+y^{2} \geqslant 2 x y, x^{2}+z^{2} \geqslant 2 x z$ and $y^{2}+z^{2} \geqslant 2 y z$ and dividing by 2, we get $x^{2}+y^{2}+z^{2} \geqslant x y+y z+x z$.

$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+x z) \geqslant 3(x y+y z+x z)=36$, from which $x+y+z \geqslant 6$. Equality is achieved when $x=y=z=2$.

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