áp dụng bất đẳng thức AM-GM ta có:

1/a+1/b+1/c>=9/(a+b+c)

=> 1/a+1/b+1/c>=9/1

=> 1/a+1/b+1/c>=9

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{9}{2}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

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

Vậy \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{9}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Bài này bạn áp dụng BĐT Cô - si là ra

Áp dụng BĐT AM-GM ta có:

\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3.\sqrt[3]{abc}.\frac{3}{\sqrt[3]{abc}}=9\)

Dấu " = " xảy ra < = > a=b=c

Ta có :

\(a+b+c=1\)

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

\(\Leftrightarrow1+\dfrac{b}{a}+\dfrac{c}{a}=\dfrac{1}{a}\)(1)

Tương tự ta lại có :

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

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

Từ 1 ; 2 và 3 :

\(\left\{{}\begin{matrix}1+\dfrac{b}{a}+\dfrac{c}{a}=\dfrac{1}{a}\\1+\dfrac{a}{b}+\dfrac{c}{b}=\dfrac{1}{b}\\1+\dfrac{a}{c}+\dfrac{b}{c}=\dfrac{1}{c}\end{matrix}\right.\)

Cộng vế theo vế ta được :

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

căng thế @Dương Phan Khánh Dương

Cauchy-Schwarz: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=9\)

Nếu chưa học thì dùng bunyakovsky :

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

thử bài bất :D 

Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)

Hoàn toàn tương tự: 

\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)

\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)

Cộng (*),(**),(***) vế theo vế ta được:

\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)

Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )

Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi a=b=c=1

1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D