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

\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

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

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

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

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0

=> Hoặc a=-b hoặc b=-c hoặc c=-a

Ko mất tổng quát, g/s a=-b

a) Ta có: vì a=-b thay vào ta được:

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)

\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)

=> đpcm

b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)

=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)

oh no bài thứ nhất là dạng chứng minh cs đúng ko ,

ko thể nào là dạng tìm a,b,c đc-.-

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

hay \(\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc=abc\)

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

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

-Xét a + b = 0 => P = 2022^2021

Bạn xét tương tự với b + c = 0 và c + a = 0 dc P = 2022^2021 nhé

a+bab+a+bc(a+b+c)=0a+bab+a+bc(a+b+c)=0

(a+b)[ab+bc+ca+c2abc(a+b+c)]=0(a+b)[ab+bc+ca+c2abc(a+b+c)]=0

 ⇔ a=−b

 ⇔ b=−c

 ⇔ c=−a

Thay vào P từng cái rồi tính tiếp nhé

-Mình làm tắt được không bạn :/?

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

\(\Rightarrow\dfrac{bc+ca+ab}{abc}=\dfrac{1}{a+b+c}\)

\(\Rightarrow\left(bc+ca+ab\right)\left(a+b+c\right)=abc\)

\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc=abc\)

\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc=0\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Rightarrow a=-b\) hay \(b=-c\) hay \(c=-a\)

\(\Rightarrow c=2022\) hay \(a=2022\) hay \(b=2022\)

-Nếu \(a=-b\)\(\Rightarrow B=\dfrac{1}{a^{2021}}+\dfrac{1}{b^{2021}}+\dfrac{1}{c^{2021}}=\dfrac{1}{a^{2021}}-\dfrac{1}{a^{2021}}+\dfrac{1}{2022^{2021}}=\dfrac{1}{2022^{2021}}\)

-Tương tự các trường hợp còn lại.

\(\dfrac{a}{2021-c}+\dfrac{b}{2021-a}+\dfrac{c}{2021-b}\\ =\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ =\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)

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

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

Vì \(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\Rightarrow A.ko.phải.số.nguyên\)

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

\(=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\ge\dfrac{9}{2}-3=\dfrac{3}{2}\)

\(minP=\dfrac{3}{2}\Leftrightarrow a=b=c=\dfrac{2021}{3}\)

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

\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{ab+ac+bc}{abc}\right)=1\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)-abc=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+ac+bc\right)+c\left(ab+ac+bc\right)-abc=0\)

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

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

\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-b\\a=-c\\b=-c\end{matrix}\right.\)

Đến đây thì nghi ngờ bạn chép sai đề biểu thức R, lẽ ra phải là dấu nhân mới tính được, nếu ko thì kết quả vẫn còn 2 ẩn

\(R=\left(a^{2017}+b^{2017}\right)\left(b^{2019}+c^{2019}\right)\left(c^{2021}+a^{2021}\right)\)

Thế này mới chính xác, kết quả \(R=0\)

ko biết mk làm có đúng ko nhma có gì sai thì đừng trách mk nhé

\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{63}{a^2+b^2+c^2}\)

\(6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{a}{ac}\right)+2021\ge\dfrac{54}{ab+bc+ac}+2021\ge\dfrac{54}{a^2+b^2+c^2}+2021\)

<=>\(\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{2021}{9}\)

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

áp dụng bđt \(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)

\(p^2\le3.\left(\dfrac{1}{3\left(2a^2+b^2\right)}+\dfrac{1}{3\left(2b^2+c^2\right)}+\dfrac{1}{3\left(2c^2+a^2\right)}\right)=\dfrac{1}{2a^2+b^2}+\dfrac{1}{2b^2+c^2}+\dfrac{1}{2c^2+a^2}\)

\(< =>p^2\le\dfrac{9}{2a^2+b^2+2b^2+c^2+2c^2+a^2}\)

<=> \(p^2\le3.\dfrac{1}{a^2+b^2+c^2}=\dfrac{2021}{3}< =>p\le\sqrt{\dfrac{2021}{3}}\)

dấu bằng xảy ra khi \(a=b=c=\sqrt{\dfrac{3}{2021}}\)

\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2021\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2021\)

\(\Rightarrow2021\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\sqrt{2021.3}=\sqrt{6063}\)

Từ đó:

\(\sqrt{3\left(2a^2+b\right)}=\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}\ge\sqrt{\left(2a+b\right)^2}=2a+b\)

\(\Rightarrow\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}\le\dfrac{1}{2a+b}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\)

Tương tự: \(\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{c}+\dfrac{1}{a}\right)\)

Cộng vế:

\(\Rightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{\sqrt{6063}}{3}\)

\(P_{max}=\dfrac{\sqrt{6063}}{3}\) khi \(a=b=c=\dfrac{3}{\sqrt{6063}}\)