Lời giải:

Tìm min:

Theo BĐT AM-GM thì: $P=a^2+b^2+c^2\geq ab+bc+ac$ hay $P\geq 9$

Vậy $P_{\min}=9$. Giá trị này đạt tại $a=b=c=\sqrt{3}$

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Tìm max:

$P=a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=(a+b+c)^2-18$

Vì $a,b,c\geq 1$ nên:

$(a-1)(b-1)\geq 0\Leftrightarrow ab+1\geq a+b$

Hoàn toàn tương tự: $bc+1\geq b+c; ac+1\geq a+c$

Cộng lại: $2(a+b+c)\leq ab+bc+ac+3=12$

$\Rightarrow a+b+c\leq 6$

$\Rightarrow P=(a+b+c)^2-18\leq 6^2-18=18$

Vậy $P_{\max}=18$. Giá trị này đạt tại $(a,b,c)=(1,1,4)$ và hoán vị

\(P=\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge\dfrac{ab+bc+ca}{ab+bc+ca}=1\)

\(P_{min}=1\) khi \(a=b=c=1\)

\(P=\dfrac{\left(a+b+c\right)^2-2\left(ab+bc+ca\right)}{ab+bc+ca}=\dfrac{9}{ab+bc+ca}-2\)

Do \(a;b\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab\ge a+b-1=2-c\)

\(\Rightarrow ab+c\left(a+b\right)\ge2-c+c\left(3-c\right)=-c^2+2c+2=c\left(2-c\right)+2\ge2\)

\(\Rightarrow P\le\dfrac{9}{2}-2=\dfrac{5}{2}\)

\(P_{max}=\dfrac{5}{2}\) khi \(\left(a;b;c\right)=\left(1;2;0\right);\left(2;1;0\right)\)

Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)

Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)

Cộng vế:

\(P\ge\dfrac{a+b+c}{3}=673\)

Dấu "=" xảy ra khi \(a=b=c=673\)

\(A=\frac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)+3\)

\(=\frac{1}{2}\left(a^2-2ab+b^2\right)+\frac{1}{2}\left(b^2-2bc+c^2\right)+\frac{1}{2}\left(c^2-2ca+a^2\right)+3\)

\(=\frac{1}{2}\left(a-b\right)^2+\frac{1}{2}\left(b-c\right)^2+\frac{1}{2}\left(c-a\right)^2+3\ge3\)

\(A_{min}=3\) khi \(a=b=c=1\)

\(B=-6\left(x^2-2.\frac{5}{12}x+\frac{25}{144}\right)+\frac{25}{24}\)

\(B=-6\left(x-\frac{5}{12}\right)^2+\frac{25}{24}\ge\frac{25}{24}\)

\(B_{min}=\frac{25}{24}\) khi \(x=\frac{5}{12}\)

Lời giải:
$P=\frac{a^2b^2+b^2c^2+c^2a^2}{abc}$

Áp dụng BĐT AM-GM, dạng $(x+y+z)^2\geq 3(xy+yz+xz)$ ta có:

$(a^2b^2+b^2c^2+c^2a^2)^2\geq 3(a^2b^4c^2+a^4b^2c^2+a^2b^2c^4)$

$=3a^2b^2c^2(a^2+b^2+c^2)=3a^2b^2c^2$

$\Rightarrow a^2b^2+b^2c^2+c^2a^2\geq \sqrt{3}abc$

$\Rightarrow P=\frac{a^2b^2+b^2c^2+c^2a^2}{abc}\geq \sqrt{3}$

Vậy $P_{\min}=\sqrt{3}$. Giá trị này đạt tại $a=b=c=\frac{1}{\sqrt{3}}$

c: Ta có: \(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)

\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)

\(=a^4-2a^3b+2ab^3-b^4\)

\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)

\(=\left(a-b\right)^3\cdot\left(a+b\right)\)

đẳh A=2ab-bc-ac cộng 2012 với A còn bên kia là a2+b2+c2 +2ab-bc-âc 

Ta có \(2P=4042ca-2ab-2bc=\left(\sqrt{2021}a+\sqrt{2021}c-\dfrac{1}{\sqrt{2021}}b\right)^2-\left(2021a^2+2021c^2+\dfrac{1}{2021}b^2\right)\ge-\left(2021a^2+2021c^2+\dfrac{1}{2021}b^2\right)\).

Lại có \(2021a^2+2021c^2+\dfrac{1}{2021}b^2\le2021\left(a^2+c^2+b^2\right)=4042\).

Do đó \(2P\ge-4042\Rightarrow P\ge-2021\).

Dấu "=" xảy ra khi và chỉ khi b = 0; \(a=-c=\pm1\).

\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow ab+bc+ca=0\)

\(\Rightarrow a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)

Ta có:

\(\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^2b^2c^2}=\dfrac{3a^2b^2c^2}{a^2b^2c^2}=3\)