steppewolf wrote: Let $t_{k} = a_{1}^k + a_{2}^k +...+a_{n}^k$, where $a_{1}$, $a_{2}$, ... $a_{n}$ are positive real numbers and $k \in \mathbb{N}$. Prove that $$\frac{t_{5}^2 t_1^{6}}{15} - \frac{t_{4}^4 t_{2}^2 t_{1}^2}{6} + \frac{t_{2}^3 t_{4}^5}{10} \geq 0 $$

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The inequality can be rewriten as: $$ 2t_{5}^{2} t_{1}^{6} + 3t_{2}^3t_{4}^{5} \geq 5t_{4}^{4}t_{2}^{2}t_{1}^{2}$$ By applying $AM-GM$ in the form: $$t_{5}^{2} t_{1}^{6} + t_{5}^{2} t_{1}^{6} + t_{2}^3t_{4}^{5} + t_{2}^3t_{4}^{5} + t_{2}^3t_{4}^{5} \geq 5(t_{5}^{4} t_{4}^{15} t_{2}^{9} t_{1}^{12})^{\frac{1}{5}} $$ it remains to prove that $$t_{5}^{4} t_{4}^{15} t_{2}^{9} t_{1}^{12} \geq t_{4}^{20}t_{2}^{10}t_{1}^{10}$$which is equivalent to proving $$t_{5}^{4}t_{1}^{2} \geq t_{4}^{5} t_{2}$$ Now from Holder's inequality we get that $$(a_{1}^{5}+a_{2}^{5} + ... + a_{n}^5)^{3}(a_{1}+a_{2} + ... a_{n}) \geq (a_{1}^{4} + a_{2}^{4}+...+a_{n}^{4})^{4} $$which is equivalent to $t_{5}^{3}t_{1} \geq t_{4}^{4}$ and we easily get $t_{5}t_{1} \geq t_{4}t_{2}$ from Muirhead. By multiplying the two former inequalities we get $t_{5}^{4}t_{1}^{2} \geq t_{4}^{5} t_{2}$ which is what we needed to prove.