Let ABC be an acute-angled triangle with area | vedclass

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(C) We know that the area $R = \frac{1}{2}ab \sin C$,so $2R = ab \sin C$,which implies $4R^2 = a^2b^2 \sin^2 C$. Substituting this into the first term

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$\frac{a^2+b^2+c^2}{2}$

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(C) We know that the area $R = \frac{1}{2}ab \sin C$,so $2R = ab \sin C$,which implies $4R^2 = a^2b^2 \sin^2 C$. Substituting this into the first term: $\sqrt{a^2b^2 - 4R^2} = \sqrt{a^2b^2 - a^2b^2 \sin^2 C} = \sqrt{a^2b^2(1 - \sin^2 C)} = \sqrt{a^2b^2 \cos^2 C} = ab \cos C$. Similarly,$\sqrt{b^2c^2 - 4R^2} = bc \cos A$ and $\sqrt{c^2a^2 - 4R^2} = ca \cos B$. Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$,so $ab \cos C = \frac{a^2+b^2-c^2}{2}$. Thus,the expression becomes: $\frac{a^2+b^2-c^2}{2} + \frac{b^2+c^2-a^2}{2} + \frac{c^2+a^2-b^2}{2} = \frac{a^2+b^2+c^2}{2}$.

Let $ABC$ be an acute-angled triangle with area $R$. Then,$\sqrt{a^2 b^2-4 R^2}+\sqrt{b^2 c^2-4 R^2}+\sqrt{c^2 a^2-4 R^2} = $

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