\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)

Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)

\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)

3: \(\left\{{}\begin{matrix}a+b>=2\sqrt{ab}\\b+c>=2\sqrt{bc}\\a+c>=2\sqrt{ac}\end{matrix}\right.\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)>=8abc\)

1: =>(a+b)(a^2-ab+b^2)-ab(a+b)>=0

=>(a+b)(a^2-2ab+b^2)>=0

=>(a+b)(a-b)^2>=0(luôn đúng)

2) Áp dụng bất đẳng thức ở câu 1 ta có:

\(\dfrac{1}{a^3+b^3+abc}\le\dfrac{1}{ab\left(a+b\right)+abc}=\dfrac{1}{ab\left(a+b+c\right)}\)

Tương tự: \(\dfrac{1}{b^3+c^3+abc}\le\dfrac{1}{bc\left(a+b+c\right)}\)

và \(\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{ca\left(a+b+c\right)}\)

Cộng vế theo vế của các bất đẳng thức trên ta được:

\(\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{a+b+c}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{1}{a+b+c}.\dfrac{a+b+c}{abc}=\dfrac{1}{abc}\left(đpcm\right)\)

Dấu "=" xảy ra khi a=b=c.

Lời giải:
Bạn tham khảo cách làm tương tự tại đây:

https://hoc24.vn/cau-hoi/cho-dfracab-2017ccdfracbc-2017aadfracca-2017bbvoi-a-b-c-ne0-tinhp-left1dfracabrightleft1dfracb.161494910584

Kết quả $P=8$ hoặc $P=-1$

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

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)

\(\Rightarrow2+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)

\(\Rightarrow\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=2\)

\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

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

\(\Rightarrow a+b+c=abc\) 

Theo đề ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\)

=>\(2+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

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

=> Đpcm

có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) =2

⇒\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)² = 4

⇔\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\) =4.

⇒2 + \(\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\) =4 (do \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)=2)

⇔\(\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\) =2 

⇔ \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\) =1

⇔\(abc\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\) =abc

⇔a +b +c =abc(đpcm)

Ta có :

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

\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)

\(=2^2=2=2+2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

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

\(=a+b+c\)

\(=abc\)

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\\ \Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow2+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\\ \Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\\ \Rightarrow\dfrac{a+b+c}{abc}=1\\ \Rightarrow a+b+c=abc\left(dpcm\right)\)

Xét b²+c²-a²=(b+c)²-a²-2bc=(a+b+c)(b+c-a)-2bc=-2bc
cmtt=>P=\(\dfrac{1}{-2bc}\)+\(\dfrac{1}{-2ab}\)+\(\dfrac{1}{-2ac}\)=\(\dfrac{a+b+c}{-2abc}\)=0

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

=
\(\dfrac{a+b+c}{\left(b^2+c^2-a^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+c^2-b^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+b^2-c^2\right)\left(a+b+c\right)}\)
= 0+0+0 = 0
Vậy P= 0 
Ngu vãi ko bt đúng không nx

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

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

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

\(=\dfrac{1}{b^2+c^2-b^2-2bc-c^2}+\dfrac{1}{a^2+c^2-a^2-2ac-c^2}+\dfrac{1}{a^2+b^2-a^2-2ab-b^2}\)

\(=\dfrac{1}{-2bc}+\dfrac{1}{-2ac}+\dfrac{1}{-2ab}\)

\(=\dfrac{a}{-2bca}+\dfrac{b}{-2acb}+\dfrac{c}{-2abc}\)

\(=\dfrac{a+b+c}{-2abc}=\dfrac{0}{-2abc}=0\)