s-a-b-c-d-number-terms-n-a-b-c-d-gt-0-then-E-s-s-a-s-s-b-s-s-c-a-E-gt-n-2-b-E-gt-n-2-n-1-c-E-gt-n-n-1-d-E-gt-n-2-n-1-e-E-gt-n-2-1-

Question Number 193238 by lmcp1203 last updated on 08/Jun/23

s = a + b + c + d + \dots . .

n u m b e r t e r m s : n

{a; b; c; d \dots . .} > 0

t h e n E = s / (s - a) + s / (s - b) + s / s - c) + \dots .

a) E >= n^{2} b) E >= n^{2} / (n - 1)

c) E >= n / (n + 1) d) E >= n^{2} / (n + 1)

e) E >= n^{2} - 1

Answered by MM42 last updated on 08/Jun/23

E ⩾ n \sqrt[n]{\frac{s^{n}}{(s - a) (s - b) (s - c) \dots}} = \frac{n s}{\sqrt[n]{(s - a) (s - b) (s - c) \dots}}

⩾ \frac{n^{2} s}{(s - a) + (s - b) + (s - c) + \dots} = \frac{n^{2} s}{n s - s} = \frac{n^{2}}{n - 1} ✓ (b)

Commented by lmcp1203 last updated on 08/Jun/23

t h a n k y o u