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.
 A: This can even be shown without the help of any theorems.
Note that
$4 x y = (x+y)^2 - (x-y)^2$, likewise for the other terms.  Substituting  this into the question gives
$$
\frac{(x+y)^2 - (x-y)^2}{2 (x+y)} + \frac{(y+z)^2 - (y-z)^2}{2 (y+z)} + \frac{(x+z)^2 - (x-z)^2}{2 (x+z)} \\
 ≤ \frac12 ((x+y) + (y+z) + (x+z))
$$
or
$$
- \frac{(x-y)^2}{2 (x+y)} - \frac{(y-z)^2}{2 (y+z)}  - \frac{ (x-z)^2}{2 (x+z)}  ≤  0
$$
which is obviously true. It also shows that equality will only be obtained for $x=y=z$. $\qquad \Box$
A: 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:
$\dfrac{2xy}{x+y}+\dfrac{2yz}{y+z}+\dfrac{2xz}{x+z}\le \dfrac{2(\frac{x+y}2)^2}{x+y}+\dfrac{2(\frac{y+z}2)^2}{y+z}+\dfrac{2(\frac{x+z}2)^2}{x+z}=\dfrac{x+y}2+\dfrac{y+z}2+\dfrac{x+z}2$
