Vì 0 ≤ a ≤ b + 1 ≤ c + 2 nên ta có a + b+c ≤ (c+2)+ (c+2) + c
<=> 1 ≤ 3c+ 4 <=> -3 ≤ 3c <=> -1≤ c
Dấu bằng xảy ra <=> a+b+c=1 và a=b +1 =c+2 <=> a=1, b=0, c=1
=> Giá trị nhỏ nhất của c = -1

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là -2 đấy

\(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(\Rightarrow2.\left(a+b+c\right)=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2\sqrt{a.\frac{1}{a}}+2\sqrt{b.\frac{1}{b}}+2\sqrt{c.\frac{1}{c}}\)

                                                          \(=2+2+2=6\)

\(\Rightarrow a+b+c\ge3\)

\(P=a+b^{2019}+c^{2020}\)

   \(=a+\left(b^{2019}+1.2018\right)+\left(c^{2020}+1.2019\right)-4037\)

\(\ge a+2019.\sqrt[2019]{b^{2019}.1^{2018}}+2020.\sqrt[2020]{c^{2020}.1^{2019}}-4037\)(BDT Cauchy-Schwarz)

\(=a+2019b+2020c-4037\)

Do \(a\le b\le c\)nên

\(\Rightarrow P\ge a+2019b+2020c\)

        \(\ge a+\left(\frac{2017}{3}+\frac{4040}{3}\right)b+\left(\frac{2020}{3}+\frac{4040}{3}\right)c-4037\)

        \(\ge a+\frac{2017}{3}a+\frac{4040}{3}b+\frac{2020}{3}a+\frac{4040}{3}c-4037\)

         \(=\frac{4040}{3}.\left(a+b+c\right)-4037\)

         \(\ge4040-4037=3\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Đặt \(\left(a+1;b+1;c+1\right)=\left(x;y;z\right)\Rightarrow1\le x\le y\le z\le2\)

\(B=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}+3\) (1)

Do \(x\le y\le z\Rightarrow\left(z-y\right)\left(y-x\right)\ge0\)

\(\Leftrightarrow xy+yz\ge y^2+zx\)

\(\Leftrightarrow\dfrac{x}{z}+1\ge\dfrac{y}{z}+\dfrac{x}{y}\)

Tương tự: \(1+\dfrac{z}{x}\ge\dfrac{y}{x}+\dfrac{z}{y}\)

Cộng vế: \(2+\dfrac{x}{z}+\dfrac{z}{x}\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{y}{x}\) (2)

Từ (1); (2) \(\Rightarrow B\le2\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+5\)

Đặt \(\dfrac{z}{x}=t\Rightarrow1\le t\le2\)

\(\Rightarrow B\le2\left(t+\dfrac{1}{t}\right)+5=\dfrac{2t^2+2}{t}+5=\dfrac{2t^2+2}{t}-5+10\)

\(\Rightarrow B\le\dfrac{2t^2-5t+2}{t}+10=\dfrac{\left(t-2\right)\left(2t-1\right)}{t}+10\le10\)

\(B_{max}=10\) khi \(t=2\) hay \(\left(a;b;c\right)=\left(0;0;1\right);\left(0;1;1\right)\)

Do \(\left\{{}\begin{matrix}a\ge0\\b\ge1\\a+b+c=5\end{matrix}\right.\) \(\Rightarrow c\le4\)

\(\Rightarrow2\le c\le4\Rightarrow\left(c-2\right)\left(c-4\right)\le0\Rightarrow c^2\le6c-8\)

\(0\le a\le1< 6\Rightarrow a\left(a-6\right)\le0\Rightarrow a^2\le6a\)

\(1\le b\le2< 5\Rightarrow\left(b-1\right)\left(b-5\right)\le0\Rightarrow b^2\le6b-5\)

Cộng vế:

\(a^2+b^2+c^2\le6\left(a+b+c\right)-13=17\)

\(A_{max}=17\) khi \(\left(a;b;c\right)=\left(0;1;4\right)\)