sequence series 1

good one.
1st one is > .
i am at present not able to find if any equality will occur.

by AM GM ineq.

1+a₁≥2√a₁

1+a₂≥2√a₂
.
.
.

1+a_n≥2√a_n

adding all

S_n≥-(n-2(√a₁+√a₂+...+√a_n))
so,

it will always be greater than RHS

second part is not clearly visible,

@ dimensions....
your approach was correct (but there was a mistake in the conclusion)...

Using AM-GM,

1 + a₁ > 2√a₁
1 + a₂ > 2√a₂
.
.
.
1 + a_n > 2√a_n

Adding,

n + S_n > 2( √a₁ + √a₂ +...+ √a_n ) (Equality is not permissible bec. all a_i's cannot be equal to 1 simultaneously as 'd≠0')

\Rightarrow S_n > 2( √a₁ + √a₂ +...+ √a_n ) - n

\Rightarrow S_n > - (n - 2( √a₁ + √a₂ +...+ √a_n ))

BUT S_n < or > n - 2( √a₁ + √a₂ +...+ √a_n ) cannot be said !!!!!

what about the second one rkrish?

is my answer rite?

@ Subash...

question is not 2S_n² > n^p . √a₁a_n

It is 2S_n² > n^p . a₁ . a_n

k ill delete the above one