Ta có \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\) 

\(=\sqrt{2a\left(a+b+c\right)+\dfrac{b^2-2bc+c^2}{2}}\)

\(=\sqrt{\dfrac{4a^2+b^2+c^2+4ab+4ac-2bc}{2}}\)

\(=\sqrt{\dfrac{\left(2a+b+c\right)^2-4bc}{2}}\)

\(\le\sqrt{\dfrac{\left(2a+b+c\right)^2}{2}}\)

\(=\dfrac{2a+b+c}{\sqrt{2}}\).

Vậy \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\le\dfrac{2a+b+c}{\sqrt{2}}\). Lập 2 BĐT tương tự rồi cộng vế, ta được \(VT\le\dfrac{2a+b+c+2b+c+a+2c+a+b}{\sqrt{2}}\)

\(=\dfrac{4\left(a+b+c\right)}{\sqrt{2}}\) \(=\dfrac{4.1011}{\sqrt{2}}\) \(=2022\sqrt{2}\)

ĐTXR \(\Leftrightarrow\) \(\left\{{}\begin{matrix}ab=0\\bc=0\\ca=0\\a+b+c=1011\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(1011;0;0\right)\) hoặc các hoán vị. Vậy ta có đpcm.

\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}=2\)

\(\Rightarrow P=2+2+2=6\)

Ta có: \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)\(=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

=> a+b=2c; b+c=2a; c+a=2b

Thay vào A ta được: A=((a+b)/b)((c+b)/c)((a+c)/a)

=2c/b.2a/c.2b/a=2.2.2=8

Áp dụng t/c dtsbn:

\(\dfrac{1}{a+b}=\dfrac{2}{b+c}=\dfrac{3}{c+a}=\dfrac{1+2+3}{2\left(a+b+c\right)}=\dfrac{6}{2\left(a+b+c\right)}=\dfrac{3}{a+b+c}\)

\(\Rightarrow\left\{{}\begin{matrix}3a+3b=a+b+c\\3b+3c=2a+2b+2c\\3a+3c=3a+3b+3c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c=2a\\b=0\end{matrix}\right.\)

\(Q=\dfrac{a+2021b+c}{a+2022b+c}=\dfrac{a+2a}{a+2a}=1\)

Áp dụng t/c dtsbn ta có:

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)

\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)

\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: 

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)

Do đó: 

\(\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

Thay a+b=2c;b+c=2a và c+a=2b vào biểu thức \(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(a+b\right)}{abc}\), ta được: 

\(P=\dfrac{2a\cdot2b\cdot2c}{abc}=\dfrac{8abc}{abc}=8\)

Vậy: P=8

Ta có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\) = \(\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\) (t/c dãy tỉ số bằng nhau)

hay \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=1\) (1)

Ta cũng có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+b+c-a}{a+c}\) (t/c dãy tỉ số bằng nhau)

hay \(\dfrac{a+b-c}{c}=\dfrac{2b}{a+c}\) (2)

Từ (1) và (2) \(\Rightarrow\) \(\dfrac{2b}{a+c}=1\) \(\Leftrightarrow\) a + c = 2b (*)

Tương tự ta cũng có: a + b = 2c (**); b + c = 2a (***)

Thay (*); (**); (***) vào P ta được:

P = \(\dfrac{2a.2b.2c}{abc}\) = 2.2.2 = 8

Vậy P = 8

Chúc bn học tốt!

\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)

Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)

\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)