Let $ a,b,c$ be nonnegative real numbers. Prove that \[ a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a) \geq 0.\] （ the 2009 Entirely Legitimate Junior Mathematical Olympiad ） http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&p=1563959.) In terms of $n\ge2$, find the largest constant $c$ such that for all nonnegative $a_1,a_2,\ldots,a_n$ satisfying $a_1+a_2+\ldots+a_n=n$, the following inequality holds: \[\frac1{n+ca_1^2}+\frac1{n+ca_2^2}+\cdots+\frac1{n+ca_n^2}\le \frac{n}{n+c}.\] （ELMO Shortlist 2011, A4） http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487111&p=2729517#p2729517 Given positive reals $x,y,z$ such that $xy+yz+zx=1$, show that \[\sum_{\text{cyc}}\sqrt{(xy+kx+ky)(xz+kx+kz)}\ge k^2,\]where $k=2+\sqrt{3}$. （ELMO Shortlist 2011, A5） http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487112&p=2729518#p2729518 Let $a,b,c$ be three positive real numbers such that $ a \le b \le c$ and $a+b+c=1$. Prove that \[\frac{a+c}{\sqrt{a^2+c^2}}+\frac{b+c}{\sqrt{b^2+c^2}}+\frac{a+b}{\sqrt{a^2+b^2}} \le \frac{3\sqrt{6}(b+c)^2}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}.\] （ELMO Shortlist 2012, A2） http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2728453&sid=6827ef4989b84c04da643548d9fa2134#p2728453 http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=239&year=2012 http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=239&year=2011

Is there another solution for this nice problem ?

My solution. $x_1+x_2+x_3= 0 \implies x_1x_2= \frac{x_3^2-x_1^2-x_2^2}{2}$ $y_1+y_2+y_3= 0 \implies y_1y_2= \frac{y_3^2-y_1^2-y_2^2}{2}$ $\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}=-\frac{(x_1^2+y_1^2)+(x_2^2+y_2^2)-(x_3^2+y_3^2)}{2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}$ Set $a=\sqrt{x_1^2+y_1^2},b=\sqrt{x_2^2+y_2^2},c=\sqrt{x_3^2+y_3^2}$ We need to prove $\sum{\frac{a^2}{bc}}-\sum{\frac{a}{b}}-\sum{\frac{b}{a}}+3 \ge 0$ <-> $\frac{\sum{a^3}+3abc-\sum{a^2b+b^2a}}{abc} \ge 0$ Obviously true by schur

math154 wrote: Let $x_1,x_2,x_3,y_1,y_2,y_3$ be nonzero real numbers satisfying $x_1+x_2+x_3=0, y_1+y_2+y_3=0$. Prove that \[\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}+\frac{x_2x_3+y_2y_3}{\sqrt{(x_2^2+y_2^2)(x_3^2+y_3^2)}}+\frac{x_3x_1+y_3y_1}{\sqrt{(x_3^2+y_3^2)(x_1^2+y_1^2)}} \ge -\frac32.\] Ray Li, Max Schindler. Beautiful problem!! Let us consider three vectors $a = (x_{1} ,y_{1}) ,b = (x_{2} ,y_{2}),a = (x_{1} ,y_{1})$. See that $a+b+c =(0,0)$. So the three vectors form a triangle. Let $A,B,C$ be the angles between these vectors. We need to show that $cosA + cosB + cosC \ge \frac{-3}{2}$, where $A+B+C = 2\pi $ Let $cos\frac{A}{2} = x , cos\frac{B}{2}= y , cos\frac{C}{2} = z =>x^2 + y^2 + z^2 + 2xyz = 1$ Claim : $x^2 + y^2 +z^2 \ge\frac{3}{4} $ Proof : If $x^2 +y^2 +z^2 < \frac{3}{4} => (|xyz|)^{\frac{2}{3}} \le \frac{(x^2 +y^2 +z^2)}{3} < \frac{1}{4}=>x^2 +y^2 +z^2 +2xyz < \frac{3+1}{4} = 1$ ,contradiction. So, $cosA + cosB + cosC + 3 = 2(cos^2\frac{A}{2} + cos^2\frac{B}{2} +cos^2\frac{C}{2}) \ge \frac{3}{2}$Done!!

MonsterS wrote: $\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}=\frac{(x_1^2+y_1^2)+(x_2^2+y_2^2)-(x_3^2+y_3^2)}{2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}$ I think it should be $\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}=-\frac{(x_1^2+y_1^2)+(x_2^2+y_2^2)-(x_3^2+y_3^2)}{2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}$. Nice solution!

Thank you.I will edit it.