since $\text{area}=\frac{abc}{4R}=\frac{\sqrt{bc}(b+c)}{4}=\frac{1}{2}bc\sin A\le \frac{bc}{2}$ we obtain $\frac{b+c}{2}\le \sqrt{bc}$ hence $b=c$ and $\sin A=1\Longrightarrow \angle A=90^{\circ}$ so $\angle B=\angle C=45^{\circ}$

I have a similar solution. From sine law $\frac{a}{2\sin{A}}=R$, substituting we get $b+c=2\sin{A}\sqrt{bc}$ But from $AM-GM$ we have $b+c\geq 2\sqrt{bc}$, hence $\sin{A}=1$ or $A=90^{o}$ and moreover $b=c$, giving $B=C=45^{o}$

N.T.TUAN wrote: The circumradius $R$ of a triangle with side lengths $a, b, c$ satisfies $R =\frac{a\sqrt{bc}}{b+c}$. Find the angles of the triangle. $R =\frac{a\sqrt{bc}}{b+c}\leq\frac{a}{2}$ $2R\leq{a}$ <=>\[ A=90{o}\] \[B=C=45{o}\]

here is my solution its completely different one R=abc/4S where a,b,c denote the side lengths and S denote the area. equating the two relations we get S=(bc)^1/2(b+c)/4 and let x be the angle between the side b and c and using the formula s=bc/2 sinx (B+C)/2=(BC)^1/2 sinx so sinx =(y+1/y)/2 where y=(b/c)^1/2 sinx x cant be negative since 0<x<180 so the only place where y+1/y is less than or equal to 1 at y=1 so sinx=1 so b=c and angle between them is 90. so the angles are 45,90,45

Here is nice solution. From the sine law we get that $\frac{a}{\sin\angle BAC}=\frac{2a\sqrt{bc}}{b+c}$, after simple factorization we get that $\sin\angle BAC=\frac{(b+c)\sqrt{bc}}{2bc}$, $\cos \angle BAC=\sqrt{1-\sin^2 \angle BAC}$, $\cos \angle BAC= \sqrt{1-\frac{(b+c)^2 bc}{4b^2c^2}}= \sqrt{\frac{4b^2c^2-(b+c)^2 bc}{4b^2c^2}}= \frac{\sqrt{bc(4bc-(b+c)^2)}} {2bc}= \frac{\sqrt{bc(2bc-b^2-c^2)}} {2bc}= \frac{\sqrt{-bc(b^2-2bc+c^2)}} {2bc}= \frac{\sqrt{-bc(b-c)^2}} {2bc}$ $bc$ and $(b-c)^2$ are always greater or equal than $0$ and $-bc$ and $-bc(b-c) ^2$ will always be less than $0$ $\implies$ $(b-c)^2=0$ or $b=c$. Since $b=c$, $\cos\angle BAC=0$ and $\angle BAC=90^\circ$ and $\angle ACB=\angle ABC=45^\circ$

for storage: Consider triangle $ABC$ with sidelengths $AB=c$, $AC=b$ and $BC=a$. Let $M$ be the midpoint of $BC$. Let $O$ be the circumcentre of triangle $ABC$. Consider triangle $CMO$: Since $MO\perp BC$, we get that $$\frac{CM}{CO}=\cos{\angle MCO}\implies \cos{\angle MCO}=\frac{b+c}{2\sqrt{bc}}\geq 1,$$by AM-GM. Since $\cos$ is always $\leq 1$, we must have $\cos{\angle MCO}=0\implies \angle MCO=0^{\circ}$. Thus, $O$ lies on $BC$ and since we have equality on AM-GM, we get that $b=c$. Since $O$ lies on $BC$ and we have $OB=OC$, we get that $M=O$ and since $BC$ is the diameter of $ABC$, we get that $\triangle ABC$ is right triangle. Since $AC=b=c=AB$, then $\triangle ABC$ is isosceles right triangle. Answer. $\boxed{\triangle{ABC}\text{ is a triangle with the angles } 90^{\circ},45^{\circ},45^{\circ}.}$

nayel wrote: since $\text{area}=\frac{abc}{4R}=\frac{\sqrt{bc}(b+c)}{4}=\frac{1}{2}bc\sin A\le \frac{bc}{2}$ we obtain $\frac{b+c}{2}\le \sqrt{bc}$ hence $b=c$ and $\sin A=1\Longrightarrow \angle A=90^{\circ}$ so $\angle B=\angle C=45^{\circ}$ Very good.