Let; x=a/b,y=b/c,z=c/a,since xyz=1,this substitution is correct (with assumption wloc :ab,ac,bc>=0) your inequality becoms (a^6/a²b²c²+c²²b²/a²b²c²)(b^6/a²b²c²+a²²c²/a²b²c²)(c^6/a²b²c²+b²²a²/a²b²c²)>=(a²²c²/a²b²c²+b^3ca²/a²b²c²)(b²²a²/a²b²c²+c^3ab²/a²b²c²)(c²²b²/a²b²c²+a^3bc²/a²b²c²),equiv:(a^6+c²²b²)(b^6+a²²c²)(c^6+b²²a²)>=(a²c+b^3)(b²a+c^3)(c²b+a^3)(abc)(abc)(abc) but :(a^6+c²²b²)>=(a^3+c²b)²/2 ,& (a^3+c²b)/2>=(²root(a^3c²b)) for every triple of reals (a,b,c) /ab>=0 so :(a^6+c²²b²)>=(a^3+c²b).(²root(a^3c²b)) ,for every triple of reals (a,b,c) /ab>=0 now :(a^6+c²²b²)(b^6+a²²c²)(c^6+b²²a²)>=(a²c+b^3)(b²a+c^3)(c²b+a^3)(²root(a^3c²b.b^3a²c.c^3b²a))=(a²c+b^3)(b²a+c^3)(c²b+a^3)(abc)(abc)(abc)&you are done!

we have by caushy schwartz $(x^4+\frac{z^2}{y^2})(y^4z^4+\frac{z^2y^2}{x^2})\geq(\frac{z^2}{x}+1)^2 $ and cyclicly thus $8LHS\geq(RHS)^2$ so we need to prove $RHS\geq8$ wich is true by am gm

Another way to solve it is to multiply everything out and you're left with: $\sum_{sym} x^6y^2 \ge \sum_{sym}x^3y$ Now we have that $2LHS = 2\sum_{sym} x^6y^2 \ge \sum_{sym} x^6y^2 + 6 \ge 2RHS$, where the last inequality is true as the summands can be rearranged into $\sum_{sym} (x^3y - 1)^2$

Stefan4024 wrote: Let $x,y,z \in \mathbb{R}$ such that $xyz = 1$. Prove that: $$\left(x^4 + \frac{z^2}{y^2}\right)\left(y^4 + \frac{x^2}{z^2}\right)\left(z^4 + \frac{y^2}{x^2}\right) \ge \left(\frac{x^2}{y} + 1 \right)\left(\frac{y^2}{z} + 1 \right)\left(\frac{z^2}{x} + 1 \right)$$ Let $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$,then the inequality becomes that \[(\frac{a^4}{b^4}+\frac{c^4}{a^2b^2})(\frac{b^4}{c^4}+\frac{a^4}{b^2c^2})(\frac{c^4}{a^4}+\frac{b^4}{c^2a^2})\ge{(\frac{a^2c}{b^3}+1)(\frac{b^2a}{c^3}+1)(\frac{c^2b}{a^3}+1)}\]By computer,we have \[LHS-RHS=\frac{1}{2}\sum_{cyc}{\frac{(a^2b^4+c^6)(a^5-b^4c)^2}{c^4b^6a^6}}\ge{0}\]

Stefan4024 wrote: Let $x,y,z \in \mathbb{R}$ such that $xyz = 1$. Prove that: $$\left(x^4 + \frac{z^2}{y^2}\right)\left(y^4 + \frac{x^2}{z^2}\right)\left(z^4 + \frac{y^2}{x^2}\right) \ge \left(\frac{x^2}{y} + 1 \right)\left(\frac{y^2}{z} + 1 \right)\left(\frac{z^2}{x} + 1 \right)$$ Proof of Lijianping:

Attachments:

very inteligent!

sqing wrote: Stefan4024 wrote: Let $x,y,z \in \mathbb{R}$ such that $xyz = 1$. Prove that: $$\left(x^4 + \frac{z^2}{y^2}\right)\left(y^4 + \frac{x^2}{z^2}\right)\left(z^4 + \frac{y^2}{x^2}\right) \ge \left(\frac{x^2}{y} + 1 \right)\left(\frac{y^2}{z} + 1 \right)\left(\frac{z^2}{x} + 1 \right)$$ Proof of Lijianping: nice proof

Let $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$.Then we have to prove $a^{8}b^{10}+a^{10}c^{8}+a^{12}b^{4}c^{2}+b^{8}c^{10}+a^{2}b^{12}c^{4}+a^{4}b^{2}c^{12}$ $\geq$ $a^{9}b^{5}c^{4}+a^{5}b^{4}c^{9}+a^{8}b^{3}c^{7}+a^{4}b^{9}c^{5}+a^{7}b^{8}c^{3}+a^{3}b^{7}c^{8}$ This Obviously holds by AM-GM inequality

can we use Ravii Substitution here?

Giffunk wrote: can we use Ravii Substitution here? Ravi substitution is for sidelength of triangles