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# Interesting Algebra Question

Note by Saran Balachandar
1 year, 10 months ago

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Using the first equation and AM - GM inequality, we have

$$\frac{a+b}{2} \geq \sqrt{ab}$$

$$4 \geq ab$$

From the second eq.

$$c^2 - 2\sqrt{3}c + 3 = \frac{ab}{2}- 2$$

$$(c - \sqrt{3})^2 = \frac{ab}{2} - 2$$

Since the left side is a square and thus positive, the right side must be positive, that is we must have $$\frac{ab}{2} \geq 2 \implies ab \geq 4$$

Therefore, we see that $$ab = 4$$. Which implies $$a = b = 2$$ and $$c = \sqrt{3}$$ · 1 year, 10 months ago

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