Prove that $3\left(\frac{x}{y} + \frac{y}{z} + \frac{z}{x}\right) \geq 2(x + y + z)\left (\frac{1}{y + z} + \frac{1}{x + z} + \frac{1}{x + y}\right)$

Question

Let $x, y, z > 0$. Prove that:

$$3\left(\frac{x}{y} + \frac{y}{z} + \frac{z}{x}\right) \geq 2(x + y + z) \left(\frac{1}{y + z} + \frac{1}{x + z} + \frac{1}{x + y}\right)$$

I tried proving this using the substitutions: $a = \frac{x}{y}, b = \frac{y}{z}, c = \frac{z}{x}$ and then using the inequality between hamonic and arithmetic mean and also some other substitutions, but nothing worked.

How can we prove this inequality?

Answer 1

By C-S $$\sum_{cyc}\frac{x}{y}=\sum_{cyc}\frac{x^2}{xy}\geq\frac{(x+y+z)^2}{xy+xz+yz}.$$ Thus, it's enough to prove that $$\frac{3(x+y+z)}{xy+xz+yz}\geq2\sum_{cyc}\frac{1}{x+y}$$ or $$\sum_{cyc}(x^3y+x^3z-2x^2yz)\geq0$$ or $$\sum_{cyc}(x^3z+y^3z-x^2yz-y^2xz)\geq0$$ or $$\sum_{cyc}z(x+y)(x-y)^2\geq0.$$ Also, by AM-GM $$\sum_{cyc}(x^3y+x^3z-2x^2yz)=\sum_{cyc}(x^3y+y^3x-2x^2yz)\geq$$ $$\geq\sum_{cyc}(2x^2y^2-2x^2yz)=\sum_{cyc}z^2(x-y)^2\geq0.$$

Answer 2

Another proof:

Use this result: $$\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$$

With $a=x,\,b=y,\,c=z$. We prove this stronger:

$$3\cdot (\frac{3}{2} +\sum\limits_{cyc}\frac{x}{y+z}) \geqq \Big[\sum\limits_{cyc} (x+y)\Big]\cdot(\sum\limits_{cyc}\frac{1}{x+y})$$

Or $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqq \frac{3}{2}$$

I think now you can end it very easy by SOS or C-S (Titu's Lemma) or Nesbitt!