Good problem

parmenides51 wrote: Let $I$ be the point of intersection of the angle bisectors of the $\vartriangle ABC$, $W_1,W_2,W_3$ be point of intersection of lines $AI, BI, CI$ with the circle circumscribed around the triangle, $r$ and $R$ be the radii of inscribed and circumscribed circles respectively. Prove the inequality $$IW_1+ IW_2 + IW_3> 2R + \sqrt{2Rr.}$$ It's wrong. Even if you mean $IW_1+ IW_2 + IW_3\geq2R + \sqrt{2Rr},$ it's wrong. Even the following is wrong:$IW_1+ IW_2 + IW_3\geq3\sqrt{2Rr}.$ PS. It was delirium...

problem comes from page 30 here (it needs $\ge$ and not $>$ in the inequality)

maybe there is a typo, or my translation is incorrect (original wording

)

Attachments:

Your translation is right. PS: The following is my delirium. The problem is not right. There is a problem for $a=b$ and $c\rightarrow0^+$.

Dear parmenides5. I found a mistake in my solution: $IW_1=CW_1$ and not $IW_1=IC$. I am sorry.

Yes, it's true. Since $IW_1=CW_1=2R\sin\frac{\alpha}{2},$ we need to prove that: $$2R\sum_{cyc}\sin\frac{\alpha}{2}\geq2R+\sqrt{2Rr}$$or $$\sqrt{2R}\sum_{cyc}\sin\frac{\alpha}{2}\geq\sqrt{2R}+\sqrt{r}$$or after Ravi's substitution $$\sum_{cyc}\sqrt{xy(x+y)}\geq\sqrt{\prod_{cyc}(x+y)}+\sqrt{2xyz}$$or after squaring of the both sides $$\sum_{cyc}\sqrt{x(x+y)(x+z)}\geq2\sqrt{xyz}+\sqrt{2\prod_{cyc}(x+y)}.$$Now, let $\sum_{cyc}(x^2y+x^2z)=6txyz$. Thus, $t\geq1$ and by C-S and Schur we obtain: $$\sum_{cyc}\sqrt{x(x+y)(x+z)}=\sqrt{\sum_{cyc}(x^3+x^2y+x^2z+xyz+2\sqrt{(x^2(x+y+z)+xyz)(y^2(x+y+z)+xyz)}}\geq$$$$\geq\sqrt{\sum_{cyc}(2x^2y+2x^2z+2xy(x+y+z)+2xyz)}=2\sqrt{xyz(6t+3)}.$$Id est, it's enough to prove that: $$2\sqrt{6t+3}\geq2+\sqrt{2(6t+2)}$$or $$6t+3\geq1+2\sqrt{3t+1}+3t+1$$or $$\sqrt{3t+1}\geq2$$and we are done!

Good one.