[不等式] 正数$x+y+z=1$证$\sum(2x-x^2)/(2x^2-2x+1)\leqslant3$

已知 $x, y, z>0$ 且 $x+y+z=1 $，求证
\[\frac{2x- x^2}{2x^2-2x+1}+\frac{2y- y^2}{2y^2-2y+1}+\frac{2z- z^2}{2z^2-2z+1}\leqslant3.\]

1# kuing


这个不等式看起来简单，可做不起来。
看来，我还得不断学习啊。

2# 我为中华添光彩


我也表示不会用Cauchy-Schwarz证。但SOS还是可以的。。

3# pxchg1200


嗯，懒一点的话用 Schur 分拆（齐次化之后）也可以

4# kuing


kk，你说有这个么？
\[   \frac{2x-x^{2}}{2x^{2}-2x+1}\leq 3x \]

5# pxchg1200

有的话我就不发上来哩

6# kuing


好吧，看见Can 贴解答了。转载下：
We will consider two cases:

• Case $\min\{x,\,y,\, z \} \ge \frac{1}{4}.$ It is easy to check that for any $a \ge \frac{1}{4},$ we have \[\frac{2a-a^2}{2a^2-2a+1} \le \frac{18a-1}{5}.\] (After expanding, it is equivalent to $(4a-1)(3a-1)^2 \ge 0,$ which is obvious.) So, \[ \frac{2x-x^{2}}{2x^{2}-2x+1}+\frac{2y-y^{2}}{2y^{2}-2y+1}+\frac{2z-z^{2}}{2z^{2}-2z+1} \le \frac{18x-1}{5}+\frac{18y-1}{5}+\frac{18z-1}{5} =3.\]
  
• Case $\min\{x,\,y,\,z\} <\frac{1}{4}.$ Without loss of generality, assume that $x<\frac{1}{4}.$ Since $2y-y^2 \ge 0$ and $2y^2-2y+1 =2\left(y-\frac{1}{2}\right)^2+\frac{1}{2} \ge \frac{1}{2},$ we have \[\frac{2y-y^2}{2y^2-2y+1} \le 2(2y-y^2).\] Similarly, we also have \[\frac{2z-z^2}{2z^2-2z+1} \le 2(2z-z^2).\] Using these two inequalities in combination with the Cauchy-Schwarz inequality, we get \[\begin{aligned} \frac{2y-y^2}{2y^2-2y+1}+\frac{2z-z^2}{2z^2-2z+1}& \le 4(y+z) -2(y^2+z^2) \le 4(y+z)-(y+z)^2\\ &=4(1-x)-(1-x)^2=3-2x-x^2.\end{aligned}\] Therefore, it suffices to prove that \[\frac{2x-x^2}{2x^2-2x+1} \le 2x+x^2.\] After expanding, it becomes \[x^2(1-x-x^2) \ge 0,\] which is true because $x^2+x< \frac{1}{16}+\frac{1}{4}<1.$

The proof is completed. $\blacksquare$

原来是分类讨论用切线法
我当时也想过，可惜<1/4时证不出来

http://www.artofproblemsolving.c ... p?f=52&t=458507

这贴里好像说上面那个证法错了

For the first one, I notice that (it is easy to find this identity)
\[ {\sum\frac{2x-x^{2}}{2x^{2}-2x+1}-3 =\frac{4(x-y)^{2}(y-z)^{2}(z-x)^{2}}{\prod (2x^{2}-2x+1)}-\sum\frac{(y-z)^{2}(y+z-x)^{2}}{(2y^{2}-2y+1)(2z^{2}-2z+1)},} \]

好一个 easy to find …………

10# kuing


他说他一看就知道是这个。。。
  汗。