Given three real numbers $x, y, z$ so that $1\leq x, y, z\leq 8.$ Prove that $$\sum\limits_{cyc}\frac{x}{y}\geq\sum\limits_{cyc}\frac{2x}{y+ z}$$

I found this inequality by discriminant, we realised the homogenous, the generality of this problem. We assume $y= 1$ that leads $x^{2}+ x^{3}- zx- zx^{4}- 3z^{2}x^{2}+ z^{2}x^{4}+ z^{3}+ z^{3}x^{3}+ z^{4}- z^{4}x\geq 0.$ By jp.Wolfram|Alpha_ https://ja.wolframalpha.com/input/?i=discriminant%5Bx%5E2%2Bx%5E3-zx-zx%5E4-3z%5E2x%5E2%2Bz%5E2x%5E4%2Bz%5E3%2Bz%5E3x%5E3%2Bz%5E4-z%5E4x%2Cx%5D%3C%3D0 _of course $constant= 8$ is not the best here I think the one way that we can deal it is using substitution in $x, y, z$ but inhomogenous. I need to the help, thanks a real lot


Solution 1:

Eliminating denominators, we arrive to the inequality $\;f(x,y,z)\ge 0,\;$ where $$\begin{align} &f(x,y,z) = x^3 y^3 + y^3 z^3 + x^3 z^3 - 3 x^2 y^2 z^2 \\[4pt]& + x^2 y^4 + y^2 z^4 + z^2 x^4 - x^4 y z - x y^4 z - x y z^4. \end{align}\tag1$$ Since $(1)$ is homogenius and has the rotational symmetry, then WLOG there are two possible cases:

  • $\mathbf{1=x\le y \le z}.\;$ Assume $\;y=1+u,\; z=y+v,\;$ then $$\begin{align} &f(1,1+u,1+u+v)=u^4(v^2+v)+u^3(5v^3+11v^2+7v+2) \\[4pt]& +u^2(9v^4+30v^3+33v^2+15v+4) +u(7v^5+30v^4+44v^3+25v^2+4v) \\[4pt]& +2v^6+10v^5+18v^4+14v^3+4v^2 \ge0, \end{align}$$ and the case is proved.
  • $\mathbf{1=z\le y\le x\le 1+p}.\;$ Assume $\;y=1+pv,\; x=1+p,\;$ then

$$\begin{align} &g(p,v)=\dfrac1{p^2}f(1+p,1+pv,1) = (p^4+p^3)v^4+(p^4+7p^3+7p^2+2p)v^3 \\[4pt]& +(3p^3+12p^2+9p+4)v^2+(-p^3-p^2+p-4)v+2p+4. \end{align}\tag2$$ If $\;v=\dfrac1p,\;p\to\infty,\;$ then the given inequality does not satisfy.

If $\;\mathbf{p=7},\;$ then $$g(7,v) = 18 -389v+1684v^2+5159v^3+2744v^4 $$$$ =(18+43v)(1-12v)^2 +13(3v-14v^2)^2 + 7v^2+59v^3+196v^4 \ge 0,$$ and the case is proved.

The greatest value of $\;p\;$ can be defined from the system $\;g'_v =g=0,\;$ with the numerical solution $\;p\approx7.07031\,16307,\; v\approx 0.08248\,18190.\;$

The best constant is 8.07031 16307.