polynomial with n real roots

clearly all roots are negative

so for positive integers x_i, i = 1,2,3 .....n

P(x) = (x+a₁)(x+a₂)....(x+a_n)

P(1) = (1+a₁)(1+a₂) ...(1+a_n)

P(1)≥2ⁿ⁺¹ from AM-GM

if a id a root of f(x) (x-a) is a factor of f(x)

i m still unable to get the way u applies am-gm!plz show that last step of applying am-gm[sorry for being dumb]

No , actually i should have shown it

(1+ a₁)≥2√a₁ (by AM - GM)

(1+ a₂)≥2√a₂

.......

(1+a_n) ≥2√a_n

Multiplying all of them
we get p(1) ≥ 2ⁿ√(a₁a₂.....a_n)= 2ⁿ⁺¹

because (a₁a₂.....a_n) = 4