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Prove that if $0\lt a\lt b$ , then $a^2\lt \frac{1}{3}\left(a^2+ab+b^2\right)\lt b^2$ .

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Lesson Parent:

Rules of Inequalities

Problem:

Prove that if $0\lt a\lt b$ , then $a^2\lt \frac{1}{3}\left(a^2+ab+b^2\right)\lt b^2$ .

Answer:

It is true that if $0\lt a\lt b$ , then $a^2\lt \frac{1}{3}\left(a^2+ab+b^2\right)\lt b^2$ .

Solution:

First, note that from this problem we know that <Sign in to see all the formulas>. Using our Rules of Inequalities we have

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Now for some more Algebra:

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and

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Putting it all together we see that

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