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I would like to prove the statement:

If $|a| > |b|$, with $a > 0$, where $a$ and $b$ are real numbers, then $|a + a^{2}| > |b + b^{2}|$.

I am fairly certain that this claim is true. Originally I did not have the restriction that $a$ must be positive, but it was demonstrated that this is a case where $|a| > |b|$ implies $|a + a^{2}| > |b + b^{2}|$. More information about this can be seen here and here.

The only way I can think of solving this is by doing a "proof by exhaustion of cases", which does not seem very fun. I am looking for a more elegant way of proving it. A more elegant proof might include use of the triangle inequality, or something else I have not thought about. If no elegant proof can be demonstrated, I will be satisfied with a "proof by exhaustion of cases".

I greatly appreciate any response.

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$|b+b^2|\leq |b|+|b|^2<a+a^2=|a+a^2|$

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  • $\begingroup$ Oh, that works. I did not see this because I incorrectly remembered the triangle inequality the opposite way. I am not sure which is more embarrassing to admit. $\endgroup$ – ddddDDDD Feb 25 '16 at 19:41

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