Minimum value of the sum (part 2)

Algebra Level pending

There are n n positive real numbers a 1 , a 2 , a 3 , . . . , a n a_1,a_2,a_3,...,a_n . Will the following inequality hold always?

a 1 n + 1 + a 2 n + 1 + a 3 n + 1 + . . . + a n n + 1 ≥ a 1 a 2 a 3 . . . a n ( a 1 + a 2 + a 3 + . . . + a n ) a_1^{n+1}+a_2^{n+1}+a_3^{n+1}+...+a_n^{n+1}\geq a_1a_2a_3...a_n(a_1+a_2+a_3+...+a_n)

No Yes

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1 solution

Nitin Kumar
May 8, 2020

By Power-Mean inequality, we have
a 1 n + 1 + a 2 n + 1 + a 3 n + 1 + . . . + a n n + 1 n ≥ ( a 1 + a 2 + a 3 + . . . + a n n ) n + 1 = ( a 1 + a 2 + a 3 + . . . + a n n ) n ( a 1 + a 2 + a 3 + . . . + a n n ) ≥ [ ( a 1 a 2 a 3 . . . a n ) 1 / n ] n ( a 1 + a 2 + a 3 + . . . + a n ) = a 1 a 2 a 3 . . . . . . a n ( a 1 + a 2 + a 3 + . . . + a n ) \frac{a_1^{n+1}+a_2^{n+1}+a_3^{n+1}+...+a_n^{n+1}}{n} \ge (\frac{a_1+a_2+a_3+...+a_n}{n})^{n+1}= (\frac{a_1+a_2+a_3+...+a_n}{n})^{n}(\frac{a_1+a_2+a_3+...+a_n}{n}) \ge [(a_1a_2a_3...a_n)^{1/n}]^n(a_1+a_2+a_3+...+a_n)=a_1a_2a_3......a_n(a_1+a_2+a_3+...+a_n)
Hence, the given inequality will always hold.
note: AM-GM inequality was used between the 3rd and 4th parts of the above inequality.

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