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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