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Prove an inequality by induction

Consider the inequality

    \[ (1+x)^n > 1+nx+nx^2 \qquad \text{for } x > 0. \]

Find the smallest integer for which this inequality holds, and prove that it holds for all larger integers.

Claim: The inequality

    \[ (1+x)^n > 1+nx+nx^2 \]

for all n \geq 3 (and for no smaller positive integers n).
Proof. First, we show that it is not true for n=1 or n=2. If n=1, then we have

    \[ (1+x) \not> 1+x+x^2 \]

since x>0 implies x^2 > 0, so the right side is larger than the left.
If n=2, then we have

    \[ (1+x)^2 = 1+2x+x^2 \not> 1+2x+2x^2 \]

since again, x^2 > 0, so 2x^2 > x^2. Therefore, the statement is false for both n=1 and n=2.
Now, for n=3 we have

    \[ (1+x)^3 = 1+3x+3x^2+x^3 > 1+3x+3x^2 \]

since x^3 > 0. Hence, the statement is true for n=3.
Now assume the statement is true for some k \in \mathbb{Z}_{\geq 3}.` So, we have,

    \begin{align*}  &&(1+x)^k &> 1+kx+kx^2 & (\text{Ind. Hyp.})\\ \implies && (1+x)^k (1+x) &> (1+kx+kx^2)(1+x) & (\text{Multiplying by } 1+x > 1)\\ \implies && (1+x)^{k+1} &> 1+kx + kx^2 + x + kx^2 + kx^3 \\ \implies && (1+x)^{k+1} &> 1+(k+1)x +(k+1)x^2. \end{align*}

Where the final inequality follows since kx^3 > 0 and 2kx^2 > (k+1)x^2, so we have just made the term on the right smaller.
Thus, if the inequality is true for some k \geq 3, then it is true for k+1. Hence, it is true for all n \geq 3. \qquad \blacksquare

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