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

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

=> (bc+ac+ab)(a+b+c)=abc

=> abc+b²c+bc²+a²c+abc+ac²+a²b+ab²+abc=abc

=>abc+b²c+bc²+a²c+abc+ac²+a²c+ab²+abc-abc=0

=>(a²c+2abc+b²c)+(a²b+ab²)+(ac²+bc²)=0

=>c(a+b)²+ab(a+b)+c²(a+b)=0

=>(a+b)[c(a+b)+ab+c²]=0

=>(a+b)(ac+bc+ab+c²)=0

=>(a+b)[a(c+b)+c(b+c)]=0

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

=> a+b=0, c+b=0, a+c=0

nếu a+b=0=>a=-b

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

và \(\dfrac{1}{a^3+b^3+c^3}=\dfrac{1}{-b^3+b^3+c^3}=\dfrac{1}{c^3}\) (2)

từ (1) và (2) suy ra đpcm

\(\dfrac{1}{a^3}+a\ge2\sqrt{\dfrac{a}{a^3}}=\dfrac{2}{a}\) ; \(\dfrac{1}{b^3}+b\ge\dfrac{2}{b}\) ; \(\dfrac{1}{c^3}+c\ge\dfrac{2}{c}\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (1)

Lại có \(\dfrac{4a}{a^4+1}\le\dfrac{4a}{2\sqrt{a^4}}=\dfrac{4a}{2a^2}=\dfrac{2}{a}\)

Tương tự \(\dfrac{4b}{b^4+1}\le\dfrac{2}{b}\) ; \(\dfrac{4c}{c^4+1}\le\dfrac{2}{c}\)

\(\Rightarrow4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\le2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (2)

Từ (1),(2)\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\)

Dấu "=" xảy ra khi a=b=c=1

Đặt vế trái BĐT cần chứng minh là P

Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ( Tự chứng minh BĐT này ), ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\le\dfrac{1}{\dfrac{4}{a+b}}=\dfrac{a+b}{4}\left(1\right)\)

Tương tự: \(\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}\le\dfrac{b+c}{4}\left(2\right)\)

\(\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\le\dfrac{c+a}{4}\left(3\right)\)

Cộng \(\left(1\right),\left(2\right),\left(3\right)\) vế theo vế, ta được:

\(P\le\dfrac{a+b+b+c+c+a}{4}=\dfrac{a+b+c}{2}\)

Dấu ''='' xảy ra khi và chỉ khi a=b=c

\(a+b+c=3abc\Rightarrow\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{ab}=3\)

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

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=9\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\cdot3=9\)

Vậy \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\).

Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)

\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)

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

\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)

\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).

ĐTXR \(\Leftrightarrow a=b=c=1\)