~~~~~~~~~~T=\frac{4}{1-a}-\frac{1}{a}+\frac{4}{1-b}-\frac{1}{b}+\frac{4}{1-c}-\frac{1}{c}
=\frac{5a-1}{a-a^2}+\frac{5b-1}{b-b^2}+\frac{5c-1}{c-c^2}
Dự đoán dấu = xảy ra khi a=b=c=\frac{1}{3}
\frac{5a-1}{a-a^2} \le ma+n
<=> 5a-1 \le ma^2 +na-ma^3-na^2
<=> ma^3+a^2(n-m)+a(5-n)-1 \le 0
Xét f(a)=ma^3+a^2(n-m)+a(5-n)-1
f'(a)=3a^2m+2a(n-m+5-n
Dấu = xảy ra khi a=\frac{1}{3}
=> \begin{cases}f(\frac{1}{3})=0 \\ f'(\frac{1}{3})=0 \end{cases}
=> \begin{cases}\frac{2}{27}m+\frac{2}{9}n=\frac{2}{3} \\ \frac{1}{3}m+\frac{1}{3}n=5 \end{cases}
=> \begin{cases}m=18 \\ n=-3 \end{cases}
~~~~~~~~~~
Ta chứng minh: \frac{5a-1}{a-a^2}\le 18a-3