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

<=>\(\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)

<=>c(a+b)(a+b+c)=-ab(a+b)

<=>(a+b)(ac+bc+c²)+ab(a+b)=0

<=>(a+b)(ac+bc+ab+c²)=0

<=>(a+b)(a+c)(c+b)=0

       a+b=0

<=> b+c=o

       c+a=0
 

a)Áp dụng bđt AM-GM cho 6 số không âm a+b,b+c,c+a ta được

\(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

TT\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

Nhân vế theo vế ta được:\(2\left(a+b+c\right)\cdot\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\)\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(đpcm\right)\)

bài toán cực trị có ẩn trong đoạn là pahir cẩn thận này @
\(0\le a,b,c\le1\)\(\Rightarrow a\left(1-a\right)\left(1-b\right)\ge0\Leftrightarrow a-ab-a^2+a^2b\ge0\)
\(\Leftrightarrow a^2b\ge ab+a^2-a\)
Tương tự \(b^2c\ge bc+b^2-b;c^2a\ge ca+c^2-c\)
\(\Rightarrow a^2b+b^2c+c^2a+1\ge1+bc+ca+ab-a-b-c+a^2+b^2+c^2\)
\(\ge\left(1-a\right)\left(1-b\right)\left(1-c\right)+abc+a^2+b^2+c^2\ge a^2+b^2+c^2\)
dấu = xảy ra \(\Leftrightarrow\left(a,b,c\right)\in\hept{ }\left(0,1,1\right),\left(0,0,1\right),\left(1,0,1\right)\left(1,1,0\right)\left(0,1,0\right),\left(1,0,0\right)\left\{\right\}\)

Do : \(\hept{\begin{cases}a\le1\Rightarrow1-a\ge0\\b\le1\Rightarrow1-b\le0\\c\le1\Rightarrow1-c\le0\end{cases}\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge0}\)

abc\(abc\ge0\Rightarrow...\)