Quote: If $a,b,c$ are the sides and $R$ the circumradius of the triangle $ABC$, prove that \[ R\ge \frac{a^2+b^2}{2\sqrt{2a^2+2b^2-c^2}}\] When does equality hold? (Baltic Way 1998) Lemma $\sin^2A\cos^2A+\sin^2B\cos^2B=\sin^2A+\sin^2B-\sin^4A-\sin^4B$

Proof of lemma

Note that $a=2R\sin A, b=2R\sin B,c=2R\sin C$. Hence $R\ge \frac{a^2+b^2}{2\sqrt{2a^2+2b^2-c^2}}$ $ \iff 2\sin^2A+2\sin^2B-\sin^2C\ge (\sin^2A+\sin^2B)^2$ Replacing $\sin C$ with $\sin (A+B)=\sin A \cos B + \cos A\sin B$, the inequality becomes $2\sin^2A+2\sin^2B\ge (\sin^2A+\sin^2B)^2+(\sin A \cos B + \cos A\sin B)^2$ Expanding and using the identity $\sin^2A+\cos^2A=\sin^2B+\cos^2B=1$ the inequality becomes $\sin^2A+\sin^2B \ge \sin^4A+\sin^4B+2\sin A \sin B \cos A \cos B$ $\iff \sin^2A+\sin^2B- \sin^4A-\sin^4B-2\sin A \sin B \cos A \cos B \ge 0$ $\iff \sin^2A\cos^2A+\sin^2B\cos^2B-2\sin A \sin B \cos A \cos B\ge 0$ $\iff(\sin A \cos A -\sin B\cos B)^2 \ge 0$ which is true. Equality occurs only when $\sin A \cos A=\sin B\cos B\implies \sin 2B=\sin 2A$, i.e. when $\angle A =\angle B$ or when $\angle C=90$. In both cases it can be checked that equality indeed occurs.

9-th Baltic Way solution

Attachments:
9Blotic Way.pdf (88kb)

$4a^2b^2c^2(2a^2+2b^2-c^2)\ge (2a^2b^2+b^2c^2+c^2a^2-(a^4+b^4+c^4))(a^2+b^2)^2\ge 0$ $(a^8+b^8-2a^4b^4)+c^4(a^4+b^4-2a^2b^2)-2c^2(a^6+b^6-a^4b^2-a^2b^4)\ge 0$ $(a^4-b^4)^2+c^4(a^2-b^2)^2-2c^2(a^2-b^2)^2(a^2+b^2)\ge 0$ $(a^2-b^2)^2(a^2+b^2-c^2)^2\ge 0$.

Also posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=146944

chinacai wrote: 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? We can rewrite this into, \[\frac{a^2b^2c^2}{2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4}\geq \frac{(a^2+b^2)^2}{4(2a^2+2b^2-c^2)};\] Or, \[\frac{c^2(2a^2+2b^2-c^2)}{2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4}\geq \frac{(a^2+b^2)^2}{4a^2b^2}.\] Subtracting $1$ from both sides we have to show that \[\frac{(a^2-b^2)^2}{2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4}\geq \frac{(a^2-b^2)^2}{4a^2b^2};\] Or, \[(a^2-b^2)^2\left[4a^2b^2-\{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\}\right];\] Which is perfectly true, being equivalent to with \[(a^2-b^2)^2(c^2-a^2-b^2)^2\geq 0.\] Equality holds iff $a=b\lor \angle A=\dfrac{\pi}{2}.\Box$ We are done. By the way, this problem also can be solved using some geometry by first rewriting this as $4Rm_a\geq a^2+b^2.$