Let $a,b,c$ the sides of a triangle, $P$ its perimeter. Prove inequality: $$\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\ge2007^3.$$

My attempt:

1) $P=a+b+c$. Then $\frac{P+2004a}{P-2a}=\frac{2005a+b+c}{b+c-a}$. Here $b+c-a>0 -$ triangle inequality.

2) $\sqrt[3]{xyz}\ge\frac{3}{\frac1x+\frac1y+\frac1z} \Rightarrow xyz\ge \left(\frac{3}{\frac1x+\frac1y+\frac1z}\right)^3$

  • $\begingroup$ @SubhadeepDey: 2007 $\endgroup$
    – Roman83
    Jun 19 '16 at 16:18

Let $$b+c-a=s,\quad c+a-b=t,\quad a+b-c=u$$ Then, we can have $$a=\frac{t+u}{2},\quad b=\frac{s+u}{2},\quad c=\frac{s+t}{2}$$

So, using AM-GM inequality and letting $d=1003$, $$\begin{align}&\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\\\\&=\left(1+d\frac ts+d\frac us\right)\left(1+d\frac st+d\frac ut\right)\left(1+d\frac su+d\frac tu\right)\\\\&=2d^3+3d^2+1+(d^3+d^2+d)\left(\frac su+\frac st+\frac tu+\frac ts+\frac us+\frac ut\right)+d^2\left(\frac{s^2}{tu}+\frac{t^2}{su}+\frac{u^2}{st}\right)\\\\&\ge 2d^3+3d^2+1+6(d^3+d^2+d)\sqrt[6]{1}+3d^2\sqrt[3]{1}\\\\&=(2d+1)^3\\\\&=2007^3\end{align}$$


I think it has to be $3\cdot 2007$ instead of $2007^3$. Since $a,b,c$ are sides of a triangle, there exist $x,y,z>0$ such that $a=y+z$, $b=z+x$ and $c=x+y$. Hence, we have to prove: $$ \sum_{cyc}\frac{2005(x+y)+(y+z)+(z+x)}{2(x+y+z)-2(x+y)}≥3\cdot 2007\iff\\ \sum_{cyc}\frac{2006x+2006y+2z}{2z}≥3\cdot 2007\iff\\ 3+1003\sum_{sym}\frac{x}{y}≥3\cdot 2007 $$ Now with AM-GM, we have $\sum_{sym}\frac{x}{y}≥6$ and thus: $$ 3+1003\sum_{sym}\frac{x}{y}≥3+6\cdot1003=3\cdot 2007 $$ Withe equality iff $x=y=z\iff a=b=c$.

  • $\begingroup$ Note LHS is product rather than sum, but I believe your method would work for the original problem as well $\endgroup$
    – Wiley
    Jun 19 '16 at 16:21

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