I solved this question with a Arithmetic mean ≥ Harmonic mean inequality. If there is another solution, please show me. $x,y,z>0$. Prove that:$$\frac{2xy}{x+y} + \frac{2yz}{y+z} + \frac{2xz}{x+z} ≤ x+y+z $$ My solution:$$\frac{x+y}{2}≥\frac{2xy}{x+y}$$ $$\frac{y+z}{2}≥\frac{2yz}{y+z}$$ $$\frac{x+z}{2}≥\frac{2xz}{x+z}$$ Summing up these inequalities $$\frac{2xy}{x+y} + \frac{2yz}{y+z} + \frac{2xz}{x+z} ≤ x+y+z $$I solved this question with a Arithmetic mean ≥ Harmonic mean inequality. If there is another solution, please show me.

Well I think using means inequalities $HM\le GM\le AM$ is pretty much the way to go, and it all boils down in the end to $(x-y)^2\ge 0$ if you decide to prove it directly.
Indeed the $AM\ge HM$ inequality is just
$AM-HM=\dfrac{x+y}2-\dfrac 2{\frac 1x+\frac 1y}=\dfrac{x+y}2-\dfrac{2xy}{x+y}=\dfrac{(x+y)^2-4xy}{x+y}=\dfrac{(x-y)^2}{x+y}\ge 0$
Even using $AM\ge GM$ inequality leads to the result:
$\dfrac{2xy}{x+y}+\dfrac{2yz}{y+z}+\dfrac{2xz}{x+z}\le \dfrac{2(\frac{x+y}2)^2}{x+y}+\dfrac{2(\frac{y+z}2)^2}{y+z}+\dfrac{2(\frac{x+z}2)^2}{x+z}=\dfrac{x+y}2+\dfrac{y+z}2+\dfrac{x+z}2$
