Quote:
Originally Posted by CaptainBlack  Let the polynomial:
f(x) = x^n + a_1 x^{n-1} + ... + a_{n-1} x +1
with non-negative real coefficients a_1, .. , a_{n-1} have n real roots.
Prove that f(2) >= 3^n
RonL |
As f(x)>0 for all x>=0 all roots are negative. Let these be -b1, -b2, .. -bn,
with all the b's >0.
Then:
f(x) =(x+b1)(x+b2) ..(x+bn)
Also as the constant term is 1 we know that b1.b2. .. bn = 1.
Now 2 + bk = 1 + 1 + bk >= 3 cuberoot(1.1.bk), by the Arithmetic-Geometric
mean inequality. So (2+bk) >= 3cuberoot(bk)
So for x=2:
f(2) = (2+b1)(2+b2) ...(2+bn) >= 3^n cuberoot(b1.b2. .. bn) = 3^n
RonL