\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)\\ =\left(ab^2-a\right)\left(c^2-1\right)+\left(a^2b-b\right)\left(c^2-1\right)+\left(a^2c-c\right)\left(b^2-1\right)\\ =ab^2c^2-ab^2-ac^2+a+a^2bc^2-a^2b-bc^2+b+a^2b^2c-a^2c-b^2c+c\\ =abc\left(ab+bc+ac\right)-\left(a^2b+ab^2+ac^2+bc^2+a^2c+b^2c\right)+\left(a+b+c\right)\\ =abc\left(ab+bc+ca\right)+\left(a+b+c\right)+3abc-\left[\left(a^2b+ab^2+abc\right)+\left(b^2c+bc^2+abc\right)+\left(a^2c+ac^2+abc\right)\right]\\ =abc\left(ab+bc+ca\right)+abc+3abc-\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\right]\\ =4abc+abc\left(ab+bc+ca\right)-\left(a+b+c\right)\left(ab+bc+ca\right)\\ =4abc+abc\left(ab+bc+ca\right)-abc\left(ab+bc+ca\right)=4abc\)

Ta có: \(a\left(b^2-1\right)\left(c^2-1\right)+b\left(c^2-1\right)\left(a^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)\)

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

\(=\left(ab^2c^2-ab^2-ac^2+a\right)+\left(bc^2a^2-bc^2-ba^2+b\right)+\left(ca^2b^2-ca^2-cb^2+c\right)\)

\(=a+b+c+ab^2c^2+bc^2a^2+ca^2b^2-ab^2-bc^2-ac^2-ba^2-ca^2-cb^2\)

\(=abc+abc.bc+abc.ca+abc.ab-ab\left(b+a\right)-bc\left(c+b\right)-ac\left(c+a\right)\)

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

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

\(=abc+abc+abc+abc=4abc\)

Vậy \(a\left(b^2-1\right)\left(c^2-1\right)+b\left(c^2-1\right)\left(a^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)=4abc\)( điều phải chứng minh ).

Lời giải:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$

$\Rightarrow (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=4$

$\Leftrightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})=4$

$\Leftrightarrow 2+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})=4$

$\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1$

$\Leftrightarrow \frac{a+b+c}{abc}=1$

$\Leftrightarrow a+b+c=abc$ (đpcm)

\(4b.ac+\left(a+c\right)^2\le4b.\dfrac{1}{4}\left(a+c\right)^2+\left(a+c\right)^2=\left(a+c\right)^2\left(b+1\right)\)

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

Lời giải:

Ta có:
\(a(b^2-1)(c^2-1)+b(a^2-1)(c^2-1)+c(a^2-1)(b^2-1)\)

\(=a(b^2c^2-b^2-c^2+1)+b(a^2c^2-a^2-c^2+1)+c(a^2b^2-a^2-b^2+1)\)

\(=(ab^2c^2+ba^2c^2+ca^2b^2)+(a+b+c)-[a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)]\)

\(=abc(ab+bc+ac)+abc-[ab(a+b)+bc(b+c)+ca(c+a)]\)

\(=abc(ab+bc+ca)+4abc-[ab(a+b+c)+bc(b+c+a)+ca(c+a+b)]\)

\(=abc(ab+bc+ca)+4abc-(a+b+c)(ab+bc+ac)\)

\(=abc(ab+bc+ca)+4abc-abc(ab+bc+ac)=4abc\)

Ta có đpcm.