安振平老师的4911号不等式问题的证明

题目：设$a,b,c\geq 0$,且$a+b+c=1$,求证：$\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}+\frac{1}{12}abc\leq \frac{7}{18}$.

证明：不妨设$a\geq b\geq c$, 令$f(b,c)=\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}+\frac{1}{12}(1-b-c)bc$, 则由已知可得$\frac{1}{3}\leq a\leq 1$, $0\leq b,c\leq 1$, 于是

$\frac{1}{48}(1-b-c)-\frac{[(b+c)^2-12(b+c)+14+2bc]}{[(b+c)^2-8(b+c)+36](b^2-4b+9)(c^2-4c+9)}\geq \frac{1}{48}(1-b-c)-\frac{[(b+c)^2-12(b+c)+14+2bc]}{36[(b+c)^2-8(b+c)+36]}$

$\geq \frac{1}{48}(1-b-c)-\frac{[\frac{3}{2}(b+c)^2-12(b+c)+14]}{36[(b+c)^2-8(b+c)+36]}=\frac{1}{48}a-\frac{[\frac{3}{2}(1-a)^2-12(1-a)+14]}{36[(1-a)^2-8(1-a)+36]}=\frac{3\left(a-\frac{1}{3}\right)^3+15\left(a-\frac{1}{3}\right)^2+60\left(a-\frac{1}{3}\right)+\frac{40}{9}}{144[(a+3)^2+20]}>0$.

所以

$f\left(\frac{b+c}{2},\frac{b+c}{2}\right)-f(b,c)=\left[\frac{1}{48}(1-b-c)-\frac{[(b+c)^2-12(b+c)+14+2bc]}{[(b+c)^2-8(b+c)+36](b^2-4b+9)(c^2-4c+9)}\right](b-c)^2\geq 0$.

从而

$\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}+\frac{1}{12}abc=\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}+\frac{1}{12}(1-b-c)bc$

$\leq \frac{1}{a^2-4a+9}+\frac{2}{\frac{1}{4}(b+c)^2-2(b+c)+9}+\frac{1}{48}(1-b-c)(b+c)^2=\frac{1}{a^2-4a+9}+\frac{2}{\frac{1}{4}(1-a)^2-2(1-a)+9}+\frac{1}{48}a(1-a)^2$.                                 (1)

又
$\frac{7}{18}-\left[\frac{1}{a^2-4a+9}+\frac{2}{\frac{1}{4}(1-a)^2-2(1-a)+9}+\frac{1}{48}a(1-a)^2\right]$

$=\frac{1-a}{144(a^2-4a+9)(a^2+6a+29)}\cdot\left[3\left(a-\frac{1}{3}\right)^6+9\left(a-\frac{1}{3}\right)^5+12\left(a-\frac{1}{3}\right)^4+\frac{24655}{162}\left(a-\frac{1}{3}\right)^2+\frac{160}{27}\left(a-\frac{1}{3}\right)+34\left(a-\frac{1}{3}\right)^2\left(a-\frac{67}{18}\right)^2+\frac{1120}{243}\right]\geq 0$.                                     (2)

由不等式(1),(2)可知原不等式成立.