Inequality x/y + y/x + z/x > (x+y+z)/xyz

[MM]

Let [tex]x,y,z>0[/tex]. Prove that [tex]\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\frac{x+y+z}{\sqrt[3]{xyz}}[/tex].

[broniran]

Since it's homogenous we can W.L.O.G assume [tex]xyz=1[/tex]. Then the inequality is equivalent to:
[tex]x^2z+y^2x+z^2y\ge x+y+z[/tex] . Now we can multiply the RHS by [tex]\sqrt[3]{(xyz)^2}=1[/tex] and now the inequality follows from the fact that [tex](2,1,0)\succ (\frac{5}{3},\frac{2}{3},\frac{2}{3})[/tex].

[MM]

now the inequality follows from the fact that [tex](2,1,0)\succ (\frac{5}{3},\frac{2}{3},\frac{2}{3})[/tex].

From where? From Muirhead's inequality? I don't think you are allowed to use it here since the inequality is cyclic and not symmetric.

[martosss]

From AM-GM we obtain
[tex]\frac{\frac{x}{y}+\frac{x}{y}+\frac{y}{z}}{3}\ge \sqrt[3]{\frac{x^2}{yz}}=\frac{x}{\sqrt[3]{xyz}}\\\frac{\frac{y}{z}+\frac{y}{z}+\frac{z}{x}}{3}\ge \sqrt[3]{\frac{y^2}{zx}}=\frac{y}{\sqrt[3]{xyz}}\\\frac{\frac{z}{x}+\frac{z}{x}+\frac{x}{y}}{3}\ge \sqrt[3]{\frac{z^2}{xy}}=\frac{z}{\sqrt[3]{xyz}}[/tex]
After summing the 3 inequalities above the solution follows.

[choisiwon]

For x, y, z are positive real numbers satisfying xyz = 1. Find Min of the expression:
[tex]P = \frac{x^{2}(y +z)}{y\sqrt{y} +2z\sqrt{z}} + \frac{y^{2}(z + x)}{z\sqrt{z} + 2x\sqrt{x}} +\frac{z^{2}(x +y)}{x\sqrt{x}+2ysqrt{y}}[/tex]

[Guest]

Thanks for solving this for me...