,![\[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1. \]](http://latex.artofproblemsolving.com/b/2/f/b2ff456632d4dd3b9ed93e4cba437bdd34d54cd8.png)
,
,
cut three common chords concurrent at 
.
,
which are internally tangent to all 
.
be a parallelogram of center 
.
,
and 
such that 
.
Y by bel.jad5, KaiRain, Adventure10
Some years ago, I saw the following problem:
Suppose
. Prove that 
.
The proof is easy. However, it is not easy to give a Sum of Squares expression of
in the form 
where 
is some integer, 
's are polynomials with non-negative coefficients, 
's are polynomials, and 
is a polynomial.
Ji Chen gave the following expression:
Results by dragonheart6:
1) Another Sum of Squares expression:
![\begin{align*}
&a^2+b^2+c^2+2abc+1 - 2(ab+bc+ca)\\
=\ & \frac{1}{2(a+b)^2}\Big[(a^2-ac-b^2+bc-a+b)^2 +(a^2-2ab-ac+b^2-bc+a+b)^2\\
&\qquad\qquad\quad + 4ab(a-b)^2 + 4ab(c-1)^2 + 4abc(a+b-2)^2\Big].
\end{align*}](//latex.artofproblemsolving.com/6/5/1/651ed700ca673fe729e7a8690d47c50dd66c5088.png)
2)
as a parameter:
i) When![$c\in [0,2]$](//latex.artofproblemsolving.com/e/0/e/e0eef817e88079b76f01b0f0e0ffe1847e607cd3.png)
,
ii) When
,
3)
as parameters:
Remark: 
for 
.
4) It is easy to prove that
for 
However, it is more difficult to find SOS.
Suppose
. Prove that 
.The proof is easy. However, it is not easy to give a Sum of Squares expression of
in the form 
where 
is some integer, 
's are polynomials with non-negative coefficients, 
's are polynomials, and 
is a polynomial.Ji Chen gave the following expression:
Results by dragonheart6:
1) Another Sum of Squares expression:
![\begin{align*}
&a^2+b^2+c^2+2abc+1 - 2(ab+bc+ca)\\
=\ & \frac{1}{2(a+b)^2}\Big[(a^2-ac-b^2+bc-a+b)^2 +(a^2-2ab-ac+b^2-bc+a+b)^2\\
&\qquad\qquad\quad + 4ab(a-b)^2 + 4ab(c-1)^2 + 4abc(a+b-2)^2\Big].
\end{align*}](http://latex.artofproblemsolving.com/6/5/1/651ed700ca673fe729e7a8690d47c50dd66c5088.png)
2)
as a parameter:i) When
![$c\in [0,2]$](http://latex.artofproblemsolving.com/e/0/e/e0eef817e88079b76f01b0f0e0ffe1847e607cd3.png)
,
ii) When
,
3)
as parameters:
Remark: 
for 
.4) It is easy to prove that
for 
However, it is more difficult to find SOS.
![$=\frac{2ab[(a+bc-b-c)^2 +(c-1)^2] +c(b-1)^2[(a+b-c)^2+1]}{2ab+c(b-1)^2}$](http://latex.artofproblemsolving.com/7/1/1/7112d77a25fc1618165d02b2e5da8c2574bf4d91.png)
and 
.
And 
Then I have:
![$$=\frac{4ab[(4c+1)(2a+4bc-1-b-c)^2 + 4c(2c-1)^2]+(4bc+b-c-1)^2[(2a+2b-c-1)^2+3(c-1)^2]}{16(4c+1)ab+4(4bc+b-c-1)^2}$$](http://latex.artofproblemsolving.com/9/6/7/967b0a1ff879b17e56fa33d75257cd05f237abce.png)
Can you check it for me?![$c \in [0,2]$](http://latex.artofproblemsolving.com/5/1/1/5118cad7693f3615a9d626fe5385fc67e39dcfda.png)
,
for 
![$=\frac{4ab[(c - 1)^2 + \frac{1}{2}c(a + b - 2)^2]+ (a - b)^2[(a + b - c)^2 + 1]}{(a + b)^2}$](http://latex.artofproblemsolving.com/2/e/a/2ea35ae0ba854a45e52a6f7df0890b21a0cf7b9a.png)
![$=\frac{3 \sum_{cyc} a(2bc-b-c+a-1)^2+ \sum_{cyc} [(b+c-a)^2+2bc+2abc+4] \cdot (b-c)^2}{(a+b+c)\ (2a+2b+2c+3)}$](http://latex.artofproblemsolving.com/d/2/8/d28e64beed9cc9dd358782add0ced02551811aac.png)
![$F(a,b,c)=\frac{3\sum_{cyc}a(a+2bc-1)^2+3\sum_{cyc} a(a-b)(a-c)+(a^2+b^2+c^2-ab-bc-ca) \cdot [(a+b+c-2)^2+4abc] }{(a+b+c)(a+b+c+3)}$](http://latex.artofproblemsolving.com/d/f/8/df87a62e86416fe05149a86fc131e3171e3d903d.png)
.
From my software:
.
![$$\,\,\,\,\,=\frac{\sum 4ab(c-1)^2 +2abc \sum (a+b-2)^2 +\sum (a-b)^2 \Big[(a+b-c)^2+1\Big]}{2(a^2+b^2+c^2+ab+bc+ca)}$$](http://latex.artofproblemsolving.com/9/c/3/9c389681cc16b5195bb8e63f0492ce475d06af6a.png)
,
,
and 
,
![$ LHS-RHS=(a-b)^2(a+b-c)+c(c-a)(c-b)+(2\sqrt[3]{abc}+1)(\sqrt[3]{abc}-1)^2+\frac{9abc(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})((\sqrt[3]{a}-\sqrt[3]{b})^2+(\sqrt[3]{b}-\sqrt[3]{c})^2+(\sqrt[3]{a}-\sqrt[3]{c})^2)}{6\sqrt[3]{abc}(a+b+c)} \geqq 0 $](http://latex.artofproblemsolving.com/d/c/a/dcace2b0f2992618ada0d2df50a48df085f25b9c.png)
,
.
You try do with 
Prove that
?![$$LHS=a^2+b^2+c^2+\frac{abc}{2}+\frac{abc}{2}+4 \ge^{AM-GM} a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \ge^{AM-GM} a^2+b^2+c^2+\frac{9abc}{a+b+c} \ge^{Schur} 2(ab+bc+ca)$$](http://latex.artofproblemsolving.com/6/2/3/623c0b578dffcd25d4ed186c2e52e588dc870d6d.png)
I fixed 
,
thus 
Case 2: 
thus 
then
