parmenides51 wrote: Given $0 < a < b < c$ prove that $$ a^{20}b^{12} + b^{20}c^{12 }+ c^{20}a^{12} <b^{20}a^{12}+ a^{20}c^{12} + c^{20}b^{12} $$ (I. Voronovich) $$\iff$$Given $0 < a < b < c$ prove that $$ a^5b^3 + b^5c^3+ c^5a^3 <b^5a^3+ a^5c^3 + c^5b^3$$

Use the following . if $x_1>x_2>...>x_n$ and $y_1>y_2>...>y_n$ and $\{t(1),t(2),...,t(n)\}=\{1,2,...,n\}$ then we have : $$max(\sum_{i=1}^n{x_iy_{f(i)}})$$happens at $t(i)=i$

sqing wrote: Given $0 < a < b < c$ prove that $$ a^5b^3 + b^5c^3+ c^5a^3 <b^5a^3+ a^5c^3 + c^5b^3$$ $$RHS-LHS=(b-a)(c-a)(c-b)\sum_{cyc}(a^2b^3+a^2c^3+2ab^2c^2+a^3bc)>0$$