If $a,b,c$ be the lengths of the sides of a triangle. Let $R$ denote its circumradius. Prove that
\[ R\ge \frac{a^2+b^2}{2\sqrt{2a^2+2b^2-c^2}}\]
When does equality hold?
1. **Given**: \(a, b, c\) are the sides of a triangle, and \(R\) is the circumradius. We need to prove:
   \[
   R \ge \frac{a^2 + b^2}{2\sqrt{2a^2 + 2b^2 - c^2}}
   \]
   and determine when equality holds.

2. **Using the Law of Sines**: 
   \[
   a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C
   \]
   Substituting these into the inequality, we get:
   \[
   R \ge \frac{(2R \sin A)^2 + (2R \sin B)^2}{2\sqrt{2(2R \sin A)^2 + 2(2R \sin B)^2 - (2R \sin C)^2}}
   \]
   Simplifying the expression:
   \[
   R \ge \frac{4R^2 (\sin^2 A + \sin^2 B)}{2\sqrt{8R^2 (\sin^2 A + \sin^2 B) - 4R^2 \sin^2 C}}
   \]
   \[
   R \ge \frac{4R^2 (\sin^2 A + \sin^2 B)}{2R \sqrt{4 (\sin^2 A + \sin^2 B) - 2 \sin^2 C}}
   \]
   \[
   R \ge \frac{2R (\sin^2 A + \sin^2 B)}{\sqrt{4 (\sin^2 A + \sin^2 B) - 2 \sin^2 C}}
   \]

3. **Simplifying the Denominator**:
   \[
   \sqrt{4 (\sin^2 A + \sin^2 B) - 2 \sin^2 C} = \sqrt{2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C)}
   \]
   Thus, the inequality becomes:
   \[
   R \ge \frac{2R (\sin^2 A + \sin^2 B)}{\sqrt{2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C)}}
   \]
   Dividing both sides by \(R\):
   \[
   1 \ge \frac{2 (\sin^2 A + \sin^2 B)}{\sqrt{2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C)}}
   \]

4. **Squaring Both Sides**:
   \[
   1 \ge \frac{4 (\sin^2 A + \sin^2 B)^2}{2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C)}
   \]
   \[
   2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C) \ge 4 (\sin^2 A + \sin^2 B)^2
   \]
   \[
   4 \sin^2 A + 4 \sin^2 B - 2 \sin^2 C \ge 4 (\sin^2 A + \sin^2 B)^2
   \]
   \[
   2 \sin^2 A + 2 \sin^2 B - \sin^2 C \ge (\sin^2 A + \sin^2 B)^2
   \]

5. **Using Trigonometric Identities**:
   \[
   \sin^2 A + \sin^2 B = 1 - \cos^2 A - \cos^2 B
   \]
   \[
   \sin^2 C = \sin^2 (A + B) = \sin^2 A \cos^2 B + \cos^2 A \sin^2 B + 2 \sin A \cos A \sin B \cos B
   \]
   Substituting \(\sin^2 C\) into the inequality:
   \[
   2 \sin^2 A + 2 \sin^2 B - (\sin^2 A \cos^2 B + \cos^2 A \sin^2 B + 2 \sin A \cos A \sin B \cos B) \ge (\sin^2 A + \sin^2 B)^2
   \]

6. **Simplifying**:
   \[
   2 \sin^2 A + 2 \sin^2 B - \sin^2 A \cos^2 B - \cos^2 A \sin^2 B - 2 \sin A \cos A \sin B \cos B \ge \sin^4 A + \sin^4 B + 2 \sin^2 A \sin^2 B
   \]
   \[
   \sin^2 A + \sin^2 B - \sin^4 A - \sin^4 B - 2 \sin A \cos A \sin B \cos B \ge 0
   \]
   \[
   (\sin A \cos A - \sin B \cos B)^2 \ge 0
   \]
   This is always true, and equality holds when \(\sin A \cos A = \sin B \cos B\), which implies \(\sin 2A = \sin 2B\). This occurs when \(\angle A = \angle B\) or \(\angle C = 90^\circ\).

The final answer is \( \boxed{ R \ge \frac{a^2 + b^2}{2\sqrt{2a^2 + 2b^2 - c^2}} } \)