By Holder's inequality , we have that $\frac{1}{3}(xy^2+yz^2+zx^2)(x^2y+y^2z+z^2x)(xy+yz+zx) \ge \Bigg(\frac{1}{\sqrt{3}} (xy + yz + zx) \Bigg)^4 \ge (x+y+z)^2(xyz)^2$ where the last inequality follows from the fact that for $a,b,c \ge 0$ , $ab + bc + ca \ge \sqrt{3abc(a+b+c)} $.

AdBondEvent wrote: By Holder's inequality , we have that $\frac{1}{3}(xy^2+yz^2+zx^2)(x^2y+y^2z+z^2x)(xy+yz+zx) \ge \Bigg(\frac{1}{\sqrt{3}} (xy + yz + zx) \Bigg)^4 \ge (x+y+z)^2(xyz)^2$ where the last inequality follows from the fact that for $a,b,c \ge 0$ , $ab + bc + ca \ge \sqrt{3abc(a+b+c)} $. Yew,AdBondEvent,you get this: $x,y,z\geq 0$,prove that \[3\, \left( {x}^{2}y+{y}^{2}z+{z}^{2}x \right) \left( x{y}^{2}+y{z}^{2 }+{x}^{2}z \right) \geq \left( xy+zx+yz \right) ^{3}\]

xzlbq wrote: AdBondEvent wrote: By Holder's inequality , we have that $\frac{1}{3}(xy^2+yz^2+zx^2)(x^2y+y^2z+z^2x)(xy+yz+zx) \ge \Bigg(\frac{1}{\sqrt{3}} (xy + yz + zx) \Bigg)^4 \ge (x+y+z)^2(xyz)^2$ where the last inequality follows from the fact that for $a,b,c \ge 0$ , $ab + bc + ca \ge \sqrt{3abc(a+b+c)} $. Yew,AdBondEvent,you get this: $x,y,z\geq 0$,prove that \[3\, \left( {x}^{2}y+{y}^{2}z+{z}^{2}x \right) \left( x{y}^{2}+y{z}^{2 }+{x}^{2}z \right) \geq \left( xy+zx+yz \right) ^{3}\] and dear,if $a,b,c$ be side of triangle,prove \[ \left( {a}^{2}b+{b}^{2}c+{c}^{2}a \right) \left( a{b}^{2}+b{c}^{2}+{ a}^{2}c \right) \geq {\frac {1}{75}}\, \left( ab+bc+ac \right) ^{2} \left( 3\,ab+3\,ac+3\,bc+22\,{b}^{2}+22\,{c}^{2}+22\,{a}^{2} \right)\] \[3\, \left( {a}^{2}b+{b}^{2}c+{c}^{2}a \right) \left( a{b}^{2}+b{c}^{2 }+{a}^{2}c \right) \geq \left( ab+bc+ac \right) ^{3}+87\, \left( b-c \right) ^{2} \left( c-a \right) ^{2} \left( a-b \right) ^{2}\] commentary： if an inequalit deos not hold,Can increase certain conditions，hlod. This is a strategy by research inequality,

Ferid.---. wrote: Let $x,y,z$ be positive real numbers.Prove that $(xy^2+yz^2+zx^2)(x^2y+y^2z+z^2x)(xy+yz+zx)\geq 3(x+y+z)^2(xyz)^2.$ JBMO Have simple proof.

Yeah. Here: $$(a^2+b^2+ab)(c^2+b^2+cb)(a^2+c^2+ac)\geq 3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)\geq (ab+bc+ca)^3$$

My proof By C-S $x^2y+y^2z+z^2x\ge \frac{(x+y+z)^2 xyz}{xy+yz+zx}$ and by AM-GM we have $xy^2+yz^2+zx^2\ge 3xyz$ Ifwe multiplie them we are done!!!

Murad.Aghazade wrote: My proof By C-S $x^2y+y^2z+z^2x\ge \frac{(x+y+z)^2 xyz}{xy+yz+zx}$ and by AM-GM we have $xy^2+yz^2+zx^2\ge 3xyz$ Ifwe multiplie them we are done!!! Very nice. Very easy.