$$\left( \dfrac{S_{BPC}}{PA^2}+ \dfrac{S_{CPA}}{PB^2}+\dfrac{S_{APB}}{PC^2} \right)\cdot( S_{BPC} \cdot PA^2+ S_{CPA} \cdot PB^2+ S_{APB} \cdot PC^2)\ge S^2$$ $$S_{BPC} \cdot PA^2+ S_{CPA} \cdot PB^2+ S_{APB} \cdot PC^2 \le S \cdot R^2$$ $\left(\dfrac{S_{BPC}}{S},\dfrac{S_{CPA}}{S},\dfrac{S_{APB}}{S}\right) - $ barycentric coordinates of point $P$

Sergic Primazon wrote: $$\left( \dfrac{S_{BPC}}{PA^2}+ \dfrac{S_{CPA}}{PB^2}+\dfrac{S_{APB}}{PC^2} \right)\cdot( S_{BPC} \cdot PA^2+ S_{CPA} \cdot PB^2+ S_{APB} \cdot PC^2)\ge S^2$$ $$S_{BPC} \cdot PA^2+ S_{CPA} \cdot PB^2+ S_{APB} \cdot PC^2 \le S \cdot R^2$$ $\left(\dfrac{S_{BPC}}{S},\dfrac{S_{CPA}}{S},\dfrac{S_{APB}}{S}\right) - $ barycentric coordinates of point $P$ Wow, I wonder why this proof is right... I just don't get it. Let's just say $ \left(\dfrac{S_{BPC}}{S},\dfrac{S_{CPA}}{S},\dfrac{S_{APB}}{S}\right)=(x,y,z) $. Then, we know that $ x+y+z=1, \sum_{cyc}^{ }x\overrightarrow{PA}=0 $. So, we know that $ \overrightarrow{PA}=y\overrightarrow{BA}+z\overrightarrow{CA}. $ That means that $ \sum_{cyc}^{ }x|\overrightarrow{PA}|^2=\sum_{cyc}^{ }(yz^2+zx^2)a^2+2xyz\sum_{cyc}^{ }\overrightarrow{BA}\cdot \overrightarrow{CA}=\sum_{cyc}^{ }yza^2 $. We now only have to show that $ \sum_{cyc}^{ }yza^2 \leq 1 $ when the radius $ R=1 $. When, radius $ R=1 $, we know that by heron, $ a^2b^2c^2=2\sum_{cyc}^{ }a^2b^2-\sum_{cyc}^{ }a^4 $ When $ x+y+z=1 $ we can show that (using lagrange multiplers or something else) that the above is maximum when $ (x,y,z)=\left ( \frac{a^2(b^2+c^2-a^2)}{\sum_{cyc}^{ }a^2(b^2+c^2-a^2)}, \frac{b^2(c^2+a^2-b^2)}{\sum_{cyc}^{ }a^2(b^2+c^2-a^2)}, \frac{c^2(a^2+b^2-c^2)}{\sum_{cyc}^{ }a^2(b^2+c^2-a^2)} \right)=\left ( \frac{b^2+c^2-a^2}{b^2c^2},\frac{c^2+a^2-b^2}{c^2a^2},\frac{a^2+b^2-c^2}{a^2b^2} \right ) $. And if we put in, we get the desired result.

Let's show this in another way. Let's show $ \left ( \sum_{cyc}^{ }yza^2 \right )\left ( \sum_{cyc}^{ }a^2(b^2+c^2-a^2) \right )\leq a^2b^2c^2(x+y+z)^2 $ We can change this into this form: $ \sum_{cyc}^{ }x^2\geq 2\sum_{cyc}^{ }yz\cos{\alpha } $ for $ \alpha +\beta +\gamma =\pi $. (It's a little dirty... ) which is very famous.

Bajisu wrote: ???

toto1234567890 wrote: Let's show this in another way. Let's show $ \left ( \sum_{cyc}^{ }yza^2 \right )\left ( \sum_{cyc}^{ }a^2(b^2+c^2-a^2) \right )\leq a^2b^2c^2(x+y+z)^2 $ We can change this into this form: $ \sum_{cyc}^{ }x^2\geq 2\sum_{cyc}^{ }yz\cos{\alpha } $ for $ \alpha +\beta +\gamma =\pi $. I think it should be $$ \sum_{cyc}^{ }x^2\geq -2\sum_{cyc}^{ }yz\cos2{\alpha }, $$but it's still true.

toto1234567890 wrote: Let's show this in another way. Let's show $ \left ( \sum_{cyc}^{ }yza \right )\left ( \sum_{cyc}^{ }a(b+c-a) \right )\leq abc(x+y+z)^2 $ We can change this into this form: $ \sum_{cyc}^{ }x^2\geq 2\sum_{cyc}^{ }yz\cos{\alpha } $ for $ \alpha +\beta +\gamma =\pi $. (It's a little dirty... ) which is very famous. Well, since it is indeed dirty, I'll post another proof. We know that the equality condition is $ (x,y,z)=\left ( \frac{a(b+c-a)}{\sum_{cyc}^{ }a(b+c-a)}, \frac{b(c+a-b)}{\sum_{cyc}^{ }a(b+c-a)}, \frac{c(a+b-c)}{\sum_{cyc}^{ }a(b+c-a)} \right)=\left ( a(b+c-a),b(c+a-b),c(a+b-c) \right ) $, so lets use this. (Let's just say that $ a^2,b^2,c^2 $s are just $a,b,c $.) Let's say that $ a'=b+c-a ,b'=c+a-b, c=a+b-c $. Then, we have to show that $ 4\left ( \sum_{cyc}^{ }yz(a'+b') \right )\left ( \sum_{cyc}^{ }a'b' \right )\leq (a'+b')(b'+c')(c'+a')(x+y+z)^2 $ Now, the equality condition has become $ (x,y,z)=(c'a'+a'b',a'b'+b'c',b'c'+c'a') $. So, we say that $ x'=y+z-x,y'=z+x-y,z'=x+y-z $. (The equality condition becomes $ (x,y,z)=(b'c',c'a',a'b') $) We have to show that $ \left ( \sum_{cyc}^{ }(x'+y')(x'+z')(b'+c') \right )\left ( \sum_{cyc}^{ }a'b' \right )\leq (a'+b')(b'+c')(c'+a')(x'+y'+z')^2 $ Let's say that $ x'=\beta\gamma, y'=\gamma\alpha, z'=\alpha\beta $ So, here it is. We have to show that $ \left ( \sum_{cyc}^{ }xy(x+z)(y+z)(a+b) \right )\left ( \sum_{cyc}^{ }ab \right )\leq (a+b)(b+c)(c+a)(xy+yz+zx)^2 $ After we expand this we get $ \sum_{cyc}^{ }x^2y^2c^2(a+b) \geq 2abcxyz\sum_{cyc}^{ }x$. This is done by AM-GM $ x^2y^2c^2a+x^2z^2b^2a \geq 2 x^2yzabc $ $ x^2y^2c^2b+y^2z^2a^2b \geq 2xy^2zabc $ $ x^2z^2b^2c+y^2z^2a^2c \geq 2xyz^2abc $