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Bounty Ended with 100 reputation awarded by Đào Thanh Oai
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Fedor Petrov
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For the first inequality, it follows (when $a_n\geqslant 1$) from Aravind's claim that $\sum_i a_i^{\alpha_i}$ is not less than arithmetic mean of $b_1,\ldots,b_n$, where $b_i=\sum_j a_i^{\alpha_j}$. Thus not less than geometric mean too.

The second inequality looks false by trivial reasons: if $a_n=0$ and all $\alpha_i$'s are positive, RHS equals 0 while LHS not necessary.

For the first inequality, it follows from Aravind's claim that $\sum_i a_i^{\alpha_i}$ is not less than arithmetic mean of $b_1,\ldots,b_n$, where $b_i=\sum_j a_i^{\alpha_j}$. Thus not less than geometric mean too.

The second inequality looks false by trivial reasons: if $a_n=0$ and all $\alpha_i$'s are positive, RHS equals 0 while LHS not necessary.

For the first inequality, it follows (when $a_n\geqslant 1$) from Aravind's claim that $\sum_i a_i^{\alpha_i}$ is not less than arithmetic mean of $b_1,\ldots,b_n$, where $b_i=\sum_j a_i^{\alpha_j}$. Thus not less than geometric mean too.

The second inequality looks false by trivial reasons: if $a_n=0$ and all $\alpha_i$'s are positive, RHS equals 0 while LHS not necessary.

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Fedor Petrov
  • 108.9k
  • 9
  • 264
  • 459

For the first inequality, it follows from Aravind's claim that $\sum_i a_i^{\alpha_i}$ is not less than arithmetic mean of $b_1,\ldots,b_n$, where $b_i=\sum_j a_i^{\alpha_j}$. Thus not less than geometric mean too.

The second inequality looks false by trivial reasons: if $a_n=0$ and all $\alpha_i$'s are positive, RHS equals 0 while LHS not necessary.