\(\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{{2}}}+{b}{x}+{c}\) are positive, and \(\displaystyle{a}+{b}+{c}={1}\) Prove

Anika Boyd

Anika Boyd

Answered question


f(x)=ax2+bx+c are positive, and a+b+c=1
Prove f(x1)f(x2)f(xn)1 is true for all positive x, satisying x1x2xn=1

Answer & Explanation

Ashley Olson

Ashley Olson

Beginner2022-03-31Added 12 answers

Step 1
It's wrong.
For n=2 the following is true.
The equality occurs for example, for
By the way, a0 and c0 by the given.
Also, for b0 we can use Holder here:

Jared Kemp

Jared Kemp

Beginner2022-04-01Added 14 answers

Step 1
(Compare A Quadratic With Positive Coefficients or Inequalities with polynomials on AoPS). Any polynomial f with real positive coefficients satisfies
for positive numbers x1,,xn. That is a consequence of the generalized Holder inequality, alternatively one can prove that ylog(f(ey)) is convex and apply Jensen's inequality.
In our case is x1xn=1 and x1xn=1, so that x1xn=1.

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