It is known $R + r = R(\cos A + \cos B + \cos C)$ so what we want to prove is $\sin^2 A + \sin^2 B + \sin^2 C \ge (\cos A + \cos B + \cos C)^2$ or, if $x^2 + y^2 + z^2 + 2xyz = 1, x^2 + y^2 + z^2 + xy + xz + yz \le \frac{3}{2}$. This is 19d in Old and New Inequalities.

If $\cos^2{A}+\cos^2{B}+\cos^2{C} \leq \frac34$ is true, then the problem is true.

shobber wrote: If $\cos^2{A}+\cos^2{B}+\cos^2{C} \leq \frac34$ is true, then the problem is true. But $\cos^2 A+\cos^2 B+\cos^2 C\leq\frac34$ is not true, exactly the opposite is true: $\cos^2 A+\cos^2 B+\cos^2 C\geq\frac34$. Here is another proof of the inequality $a^2+b^2+c^2\geq 4\left(R+r\right)^2$: As MysticTerminator noted, this inequality is equivalent to $\sin^2 A+\sin^2 B+\sin^2 C\geq\left(\cos A+\cos B+\cos C\right)^2$. Since $\sin^2 x=1-\cos^2 x$ for any angle x, this becomes $\left(1-\cos^2 A\right)+\left(1-\cos^2 B\right)+\left(1-\cos^2 C\right)\geq\left(\cos A+\cos B+\cos C\right)^2$ $\Longleftrightarrow\ \ \ 3-\left(\cos^2 A+\cos^2 B+\cos^2 C\right)\geq\left(\cos A+\cos B+\cos C\right)^2$ $\Longleftrightarrow\ \ \ \left(\cos A+\cos B+\cos C\right)^2+\left(\cos^2 A+\cos^2 B+\cos^2 C\right)\leq 3$ $\Longleftrightarrow\ \ \ \left(\cos B+\cos C\right)^2+\left(\cos C+\cos A\right)^2+\left(\cos A+\cos B\right)^2\leq 3$. But this inequality was proved on http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 . (Another of my memorial inequalities: When it was posted in July 2004, I wasn't able to solve it; all I could do was, after Fuzzylogic gave a nice solution, to rewrite it in a different way. Later, in February 2005, the same inequality was given to me by Prof. Gronau at the IMO training camp. At that time, I had absolutely forgotten about the solutions given on ML and started solving it again. After 4 hours, I came up with a solution - it was exactly the same as the one posted by Fuzzylogic...) darij

As I knowed , this ineq can be written in the form : $ p^2 \geq 2R^2 +8Rr+3r^2 $ , which is found by A.W.Walker Because $ R \geq 2r $ and $ 2R^2+10Rr-r^2+2(R-2r)\sqrt{R(R-2r)} \geq p^2 $ , so we just need to prove one in two ineqs : $ (2R+r)^2 \geq 2R^2+8Rr+3r^2 $ or $ 2R^2+10Rr-r^2+2(R-2r)\sqrt{R(R-2r)} \geq 2R^2+8Rr+3r^2 $ $ \iff R^2-2Rr-r^2 \geq 0 $ or $ R^2-2Rr-r^2 \leq 0 $ , which is true The equality holds for $ R=2r $ or $ R=(\sqrt2 +1)r $ This ineq has an equivalent form : $ (HA+HB+HC)^2 \leq AB^2+AC^2+BC^2 $ , where $ H $ is the orthocenter of triangle $ ABC $ PS : This is not my solution , it's get from my document .

Here are my solutions for this problem : First solution : I'll prove that : $ (HA+HB+HC)^2 \leq a^2+b^2+c^2 $ We have $ AH.AA' = bc.cosA = \frac{b^2+c^2-a^2}{2} $ , where $ A' $ is the feet of the altitude from $ A $ to $ BC $ So $ AH.AA' +BH.BB' +CH.CC' = \frac{a^2+b^2+c^2}{2} $ Apply Cauchy's ineq , we have : $ (HA+HB+HC)^2 = (\sum \sqrt{AH.AA'}.\sqrt{\frac{AH}{AA'}})^2 \leq \frac{a^2+b^2+c^2}{2}. \frac{AH}{AA'} = a^2+b^2+c^2 $ Second solution : We have : $ 2cosA.cosB = \sqrt{sin2A.cotA.sin2B.cotB} \leq \frac{1}{2}(sin2AcotB+sin2BcotA) $ We just need to show : $ \sum{cotA(sin2B+sin2C)= -2(cos2A+cos2B+cos2C)} = 3-2(cos^2A+cos^2B+cos^2C) $ , which is true because $ cotA(sin2B+sin2C)= -2cos(C+B).cos(C-B) = -(cos2B+cos2C) $ Now , we have over 8 solutions for this problem

MysticTerminator wrote: $\sin^2 A + \sin^2 B + \sin^2 C \ge (\cos A + \cos B + \cos C)^2$ or, if $x^2 + y^2 + z^2 + 2xyz = 1, x^2 + y^2 + z^2 + xy + xz + yz \le \frac{3}{2}$. This is 19d in Old and New Inequalities. How to prove it? Can you show?

I can only prove that $xy+yz+zx\leq\frac{3}{4}$

Remark. $4(R+r)^2\le a^2+b^2+c^2\Longleftrightarrow 2R^2+8Rr+3r^2\le s^2\ \ (1)$ $16Rr-5r^2\le 2R^2+8Rr+3r^2\Longleftrightarrow 2R^2-8Rr+8r^2\ge 0\Longleftrightarrow 2(R-2r)^2\ge 0.$ Therefore, if the triangle $ABC$ is acute, then the Blundon's inequality $16Rr-5r^2\le s^2$ is more weak than the inequality (1).

It's not Blundon's inequality. In fact this inequality is due to Gerrstern.

O.K. What is the Blundon's inequality then ?

I think $s^2\geq 16Rr-5r^2$ is Columbien -Doucet ineq(1872)

Guys, isn't it clear that some inequalities just get lots of different names? You are all talking about the same inequality, please don't worry about the name

What is the maximum value of $\cos{\alpha}+\cos{\beta}+\cos{\gamma}$ when $\alpha + \beta + \gamma =360 ^o$ I see that it is 3 ( $\alpha=0, \beta = 0 \gamma = 360$ ) But what if this are the angles between circumradiuses and sides of a triangle?

We have $\triangle ABC$ acute. Therefore, $a^2, b^2, c^2$ are sides of some triangle. Therefore, we can let $a^2 = y+z, \ b^2 = z+x, \ c^2 = x+y$ for some positive reals $x,y,z$. Also, let $x$ be the largest among $x,y,z$ so that $a$ is the smallest side, and therefore, $h_a$ the largest altitude. We have, \[ \cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{x}{\sqrt{(x+y)(x+z)}}\] and therefore, \[ \sin A = \sqrt{\frac{xy+yz+zx}{(x+y)(x+z)}} \] Thus, we have, \[ R = \frac{a}{2 \sin A} = \sqrt{\frac{(x+y)(y+z)(z+x)}{2(xy+yz+zx)}}\] Also, \[ 1+ \frac rR = \cos A + \cos B + \cos C = \frac{\sum x\sqrt{(y+z)}}{\sqrt{(x+y)(y+z)(z+x)}} \] Now, let us state the problem in terms of $x,y,z$. We have to prove, \[2(x+y+z) \ge 4(R+r)^2 = (2R)^2 (1+ \frac rR)^2 = \frac{(x+y)(y+z)(z+x)}{(xy+yz+zx)} \cdot \frac{(\sum x\sqrt{(y+z)})^2}{(x+y)(y+z)(z+x)}\] \[ \iff 2(x+y+z)(xy+yz+zx) \ge (\sum x\sqrt{(y+z)})^2 \] Now, we have by Cauchy Schwarz Inequality, $(\sum x\sqrt{(y+z)})^2 \le (\sum x(y+z)) (\sum x) = 2(xy+yz+zx)(x+y+z)$ We are through.