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Prove $3^n \geq 2n^2 +1$ for $n = 1,2,\ldots$ using induction

This is what I have so far

Base case - $n=1,3^1 \geq 2^1 + 1 = 3$ true

Induction step - Assume true for some n, then,

$ 3*3^n \geq 3*(2n^2 +1)=6n^2 +3 $

I have to somehow manipulate and show its $\geq 2*(n+1)^2 +1$

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You're almost there. $$2(n+1)^2+1 = 2(n^2+2n+1)+1 = 2n^2+4n +3 \underbrace{\leq}_{(\ast)} 2n^2+4n^2 +3 = 6n^2+3,$$ where $(\ast)$ holds since $n\geq 1$.

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First, show that this is true for $n=1$:

$3^1\geq2\cdot1^2+1$

Second, assume that this is true for $n$:

$3^n\geq2n^2+1$

Third, prove that this is true for $n+1$:

$3^{n+1}=$

$3\cdot\color\red{3^n}\geq$

$3\cdot(\color\red{2n^2+1})=$

$6n^2+3=$

$2n^2+4n^2+3\geq$

$2n^2+4n+3=$

$2n^2+4n+2+1=$

$2(n^2+2n+1)+1=$

$2(n+1)^2+1$


Please note that the assumption is used only in the part marked red.

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