I want to prove the inequality $$\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 4(\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x})$$, for positive reals. I have a hint that it is solved with Popoviciu's inequality applied to $f(x)=x+\frac{1}{x}$. I can show the the function $f$ is convex. Moreover, I see that the LHS of the inequality can be expressed as $f(\frac{x}{y})+f(\frac{x}{z})+f(\frac{z}{y})$. There are other representations as we can choose $2^3=8$ different ways to express the LHS as the sum of $f$'s. However, none of these seem to lead to the desired inequality. I considered rewriting the Popoviciu's inequality as $$f(a)+f(b)+f(c)\ge 2(f(\frac{a+b}{2})+f(\frac{a+c}{2})+f(\frac{b+c}{2}) – \frac{3}{2}f(\frac{a+b+c}{3}))$$, where $a,b,c$ equal $x/y, x/z, z/y$ respectively.
However, it doesn't seem to be the solution. After simplifying the RHS becomes: $$\frac{4}{a+b}+\frac{4}{a+c}+\frac{4}{b+c}-\frac{9}{a+b+c}+a+b+c$$.
Trying slightly different $a, b, c$, namely $x/z, y/z, y/x$ still gives the same result that I can't simplify further. Note, however, that in this case $\frac{1}{a+b}=\frac{z}{x+y}$.
I thought that maybe the RHS of my Popoviciu's inequality can be shown to be greater than or equal than the desired inequality, but I couldn't come up with anything.
Best Answer
For positives $x$, $y$ and $z$ we need to prove that $$\sum_{cyc}\frac{x+y+z}{x}-3\geq4\sum_{cyc}\frac{x+y+z}{x+y}-12$$ or $$\sum_{cyc}\frac{1}{x}+\frac{9}{x+y+z}\geq4\sum_{cyc}\frac {1}{x+y}$$ or $$\sum_{cyc}\frac{1}{x}+\frac{3}{\frac{x+y+z}{3}}\geq2\sum_{cyc}\frac {1}{\frac{x+y}{2}},$$which is Popovicui for the convex function $f(x)=\frac{1}{x}.$
There are another ways for the proof of your inequality (I see at least 10 of them).
If you want, I can show something.