Let P(x)=xn+an−1xn−1+⋯+a1x+1 where ai are nonnegative and real. Assume P has n real roots.
Prove P(2)≥3n.
I thought I had a good idea about rewriting P as (x−α1)⋯(x−αn). The fact that the roots are real means that you can order them. Choose the largest root, αk, then 2−αk would have the smallest such value among the αi. And so we would have P(2)≥(2−αk)n.
But I haven't been able to think of a way to compare it with 3n.
Answer
The proof follows from the following Lemma:
Proof: (2+a)(2+b)≥3(2+ab)⇔4+2a+2b+ab≥6+3ab⇔0≥2−2a−2b+2ab⇔0≥2(a−1)(b−1)
QED Lemma
Now, lets solve the problem. Let bi=−α1, and we can assume without loss of generality that b1≤b2≤...≤bn.
As b1⋅..⋅bn=1 it follows that bn≥1b1⋅...⋅bk≤1
Then, by repeadely applying the previous lemma, we get
(2+b1)(2+b2)(2+b3)⋅...⋅(2+bn)≥3(2+b1b2)(2+b3)⋅...⋅(2+bn)≥32(2+b1b2b3)⋅...⋅(2+bn)≥....3k−1(2+b1b2b3...bk)(2+bk+1⋅...⋅(2+bn)≥...3n−1(2+b1b2b3...bn)=3n
Much simpler second solution:
If b1,..,bn are non-negative numbers then by the AM-GM inequality:
2+bi=1+1+bi≥33√bi
Therefore (2+b1)(2+b2)(2+b3)⋅...⋅(2+bn)≥3n3√b1b2...bn=3n
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