Because all coefficients are positive so all n roots are -ve and hence
$P(x) = \prod_{k=1}^{n} (x+ a_k)$ where all $a_k$ are positive
Further $\prod_{k=1}^{n} (a_k) = 1$
So $P(2) = \prod_{k=1}^{n}(2+a_k)\cdots(1)$
Now taking AM GM between 1,1 $a_k$ we get $(2+a_k) >= 3\sqrt[3]{(a_k)}\cdots(2)$
So from (1) and (2)
$P(2) >= 3^n \sqrt[3]{\prod_{k=1}^{n} (a_k)} = 3^n$ and hence $P(2) >= 3^n$
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