\(1-c=a+b\ge2\sqrt{ab}\Rightarrow4ab\le\left(1-c\right)^2\)

\(2bc+ca\le2bc+2ca=2c\left(a+b\right)=2c\left(1-c\right)\)

Từ đó ta có:

\(P\le\left(1-c\right)^2+2c\left(1-c\right)=1-c^2\le1\)

\(P_{max}=1\) khi \(\left(a;b;c\right)=\left(\dfrac{1}{2};\dfrac{1}{2};0\right)\)

\(A=4ab+8bc+6ca=a\left(b+c\right)+3b\left(a+c\right)+5c\left(a+b\right)\)

\(=a\left(3-a\right)+3b\left(3-b\right)+5c\left(3-c\right)\)

\(=\dfrac{81}{4}-\left[\left(a-\dfrac{3}{2}\right)^2+3\left(b-\dfrac{3}{2}\right)^2+5\left(c-\dfrac{3}{2}\right)^2\right]\)

Đặt \(x=\left|a-\dfrac{3}{2}\right|;y=\left|b-\dfrac{3}{2}\right|;z=\left|c-\dfrac{3}{2}\right|\)

\(\Rightarrow x+y+z\ge\left|a+b+c-\dfrac{9}{2}\right|=\dfrac{3}{2}\)

Khi đó \(A=\dfrac{81}{4}-\left(x^2+3y^2+5z^2\right)\) 

Áp dụng bđt bunhiacopxki: \(\left(x^2+3y^2+5z^2\right)\left(\dfrac{45^2}{46^2}+\dfrac{3.15^2}{46^2}+\dfrac{5.9^2}{46^2}\right)\ge\left(\dfrac{45}{46}x+\dfrac{45}{46}y+\dfrac{45}{46}z\right)^2\ge\left(\dfrac{135}{92}\right)^2\)

\(\Leftrightarrow x^2+3y^2+5z^2\ge\dfrac{135}{92}\)

\(\Rightarrow A\le\dfrac{81}{4}-\dfrac{135}{92}=\dfrac{432}{23}\)

Dấu = xảy ra\(\Leftrightarrow x=3y=5z\) và \(x+y+z=\dfrac{3}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{45}{46}\\y=\dfrac{15}{46}\\z=\dfrac{9}{46}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{12}{23}\\b=\dfrac{27}{23}\\c=\dfrac{30}{23}\end{matrix}\right.\)

Vậy...

\(\hept{\begin{cases}a+3c=2016\\a+2b=2017\end{cases}}\left(1\right)\)

Cộng từng vế của hệ (1), ta được:

\(2a+2b+3c=4033\)

\(\Leftrightarrow2a+2b+2c=4033-c\)

\(\Leftrightarrow2\left(a+b+c\right)=4033-c\)

Vì c không âm nên \(4033-c\le4033\)

\(\Rightarrow a+b+c\le\frac{4033}{2}=2016\frac{1}{2}\)

Vậy GTLN của P là \(2016\frac{1}{2}\Leftrightarrow c=0\)

Lúc đó: \(a=2016;b=\frac{1}{2}\)

Ta có: a + 3c = 2016 ; a + 2b = 2017

Do đó : 2a + 2b + 3c = 2a + 2b + 2c + c = 2 (a + b + c) + c = 4033  

Suy ra: 2 (a + b + c) = 4033 - c

Để 2 (a + b + c) lớn nhất thì 4033 - c lớn nhất

Nên c nhỏ nhất , mà c >= 0 nên c = 0.

Từ đó ta suy ra  : 2 (a + b + c) <= 4033 <=> a + b + c <= 2016,5

Vậy Max P = 2016,5 

Khi c = 0 ; a = 2016 ; b = 0,5

Ta có M=a(b+c)+3b(c+a)+5c(a+b)=a(3-a)+3b(3-b)+5c(3-c)=\(\frac{81}{4}\)-\(\left(a-\frac{3}{2}\right)^2+3\left(b-\frac{3}{2}\right)^2+5\left(c-\frac{3}{2}\right)^2\)

Đặt x=\(\left|a-\frac{3}{2}\right|\),y=\(\left|b-\frac{3}{2}\right|\),z=\(\left|c-\frac{3}{2}\right|\)=>x+y+z\(\ge\left|a+b+c-\frac{9}{2}\right|=\frac{3}{2}\)

Khi đó M=\(\frac{81}{4}-\left(x^2+3y^2+5z^2\right)\)

Đưa thêm các tham số\(\alpha,\beta,\gamma>0\)Áp dụng bất đẳng thức AM-GM:\(x^2+\alpha^2\ge2x\alpha\)(1);\(3y^2+3\beta^2\ge6y\beta\)(2);\(5z^2+5\gamma^2\ge10z\gamma\)(3)

Suy ra: \(M-\alpha^2-3\beta^2-5\gamma^2\le\frac{81}{4}-2\left(x\alpha+3y\beta+5z\gamma\right)\)

Ta chọn \(\alpha=3\beta=5\gamma\)\(\Rightarrow M\le\frac{81}{4}+\alpha^2+3\beta^2+5\gamma^2-2\alpha\left(x+y+z\right)\)\(\le\frac{81}{4}+\alpha^2+3\beta^2+5\gamma^2-3a\)

Ta thấy dấu bằng các bất đẳng thức (1),(2),(3) xảy ra khi \(x=\alpha,y=\beta,z=\gamma\)\(\Rightarrow\alpha+\beta+\gamma=\alpha+\frac{\alpha}{3}+\frac{\alpha}{5}=x+y+z=\frac{3}{2}\)\(\Rightarrow\alpha=\frac{45}{46}\),\(\beta=\frac{15}{46},\gamma=\frac{9}{46}\)

Vậy MaxM=\(\le\frac{81}{4}+\left(\frac{45}{46}\right)^2+3\left(\frac{15}{46}\right)^2+5\left(\frac{9}{46}\right)^2-3.\frac{45}{46}\)=\(\frac{432}{23}\)

Lời giải:
Vì $a,b,c$ không âm và $a+b+c=2\Rightarrow 0\leq a,b,c\leq 2$
Khi đó:

$a\leq 12a$

$2b^2=2b.b\leq 4b\leq 12b$

$3c^3=3c^2.c\leq 3.2^2.c=12c$

$\Rightarrow P=a+2b^2+3c^3\leq 12(a+b+c)=24$
Vậy $P_{\max}=24$ khi $(a,b,c)=(0,0,2)$

a+3=8

suy ra a=5

ta có: 5+2b=9

2b=4

b=2

vậy a+b=2+5=7

tổng a+b+c</7 cóGTLN