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Note by Saran Vijai 1 year, 1 month 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 \( c = \sqrt{3} \) – Siddhartha Srivastava · 1 year, 1 month ago

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@Saran Vijai – Try the wiki. – Siddhartha Srivastava · 1 year ago

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TopNewestUsing 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 \( c = \sqrt{3} \) – Siddhartha Srivastava · 1 year, 1 month ago

Log in to reply

Log in to reply

wiki. – Siddhartha Srivastava · 1 year ago

Try theLog in to reply