Proof by induction problem explanation

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The discussion centers on the proof by induction involving the inequality (1+x)^n ≥ 1+nx and its extension to (1+x)^{n+1} ≥ 1+(n+1)x. Multiplying both sides of the initial inequality by (1+x) is justified because (1+x) is non-negative, allowing the manipulation of the inequality. The left-hand side of the resulting inequality matches the left-hand side of the target inequality, confirming the induction step. The challenge lies in verifying that the right-hand side, (1+x)(1+nx), simplifies correctly to 1+(n+1)x. The conversation emphasizes understanding the steps in the induction process and the properties of exponents.
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In this link (https://www.physicsforums.com/showthread.php?t=523874 ) there's an example of proof by induction.

Somewhere in the explanation there's the phrase:
Because x+1≥0, we can multiplicate both sides by x+1.

Why they decided to multiply both sides by x+1?
 
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It was assumed that
(1) (1+x)^n\geq 1+nx

And you want to prove that
(2) (1+x)^{n+1}\geq 1+(n+1)x

By multiplying both sides of (1) by (1+x), you'll get the LHS of the resulting inequality to match the LHS of (2).

a^m a = a^m a^1 = a^{m+1} by the product property of exponents, so
(1+x)^n (1+x) = (1+x)^{n+1}.
 
So, the equation will be: (1+x)^{n+1} \geq (1+x)(1+nx)?

From the RHS I can't get 1+(n+1)x, or can I?
 
Well, what do you get when you multiply on the right?
 
(1+x)(1+nx) is equal to 1+(n+1)x?
 
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