\(P=\frac{2a+3b+3c-1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2c+1}{2017+c}\)

\(=\frac{6047-a}{2015+a}+\frac{6048-b}{2016+b}+\frac{6049-c}{2017+c}\)

\(=\frac{8062}{2015+a}+\frac{8064}{2016+b}+\frac{8066}{2017+c}-3\)

\(\ge\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{2015+2016+2017+a+b+c}-3=\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{8064}-3\)

Dấu = xảy ra khi ....

\(M=2003\left(\dfrac{1}{a}+4a\right)+2016\left(\dfrac{1}{b}+b\right)-5012a-7518b\)

\(M=2003\left(\dfrac{1}{a}+4a\right)+2016\left(\dfrac{1}{b}+b\right)-2506\left(2a+3b\right)\)

\(M\ge2003.2\sqrt{\dfrac{4a}{a}}+2016.2\sqrt{\dfrac{b}{b}}-2506.4=2020\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{1}{2};1\right)\)

Thay a=-3b vào M 

\(DK.a\ne0;b\ne0\)

\(M_b=\frac{2a+b}{a-b}-\frac{2a-b}{a+2b}=\frac{-6b+b}{-3b-b}-\frac{-6b-b}{-3b+2b}=\frac{5}{4}-\frac{-7}{-1}=-\frac{23}{4}\)

Ta có: \(\left(a+b\right)^2\ge4ab=16\Rightarrow a+b\ge4\Rightarrow a+b-4\ge0\)

\(P=\dfrac{1+b+1+a}{\left(1+a\right)\left(1+b\right)}=\dfrac{a+b+2}{ab+a+b+1}=\dfrac{a+b+2}{a+b+5}\)

\(P=\dfrac{3a+3b+6}{3\left(a+b+5\right)}=\dfrac{2\left(a+b+5\right)+\left(a+b-4\right)}{3\left(a+b+5\right)}\ge\dfrac{2\left(a+b+5\right)}{3\left(a+b+5\right)}=\dfrac{2}{3}\)

\(P_{min}=\dfrac{2}{3}\) khi \(a=b=2\)

 Cosi: ab <= 1/4 
Quy đồng P, ta đc: 
P = (2ab+1)/(ab+2). 
Ta cm P <= 2/3 
<=> 3(2ab+1) <= 2(ab+2) 
<=> ab<= 1/4 (đúng) 
Vậy maxP = 2/3 khi a=b =1/2