Ta có \(\dfrac{2}{a-b}\)+\(\dfrac{2}{b-c}\)+\(\dfrac{2}{c-a}\)

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

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

=\(\dfrac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right).\left(a-c\right)}\)+\(\dfrac{\left(b-a\right)-\left(b-c\right)}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{\left(c-b\right)-\left(c-a\right)}{\left(c-b\right).\left(c-a\right)}\)

= \(\dfrac{a-c-a+b}{\left(a-b\right).\left(a-c\right)}\)+\(\dfrac{b-a-b+c}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{c-b-c+a}{\left(c-b\right).\left(c-a\right)}\)

= \(\dfrac{-c+b}{\left(a-b\right).\left(a-c\right)}\)+ \(\dfrac{-a+c}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{-b+a}{\left(c-b\right).\left(c-a\right)}\)

= \(\dfrac{b-c}{\left(a-b\right).\left(a-c\right)}\)+\(\dfrac{c-a}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{a-b}{\left(c-b\right).\left(c-a\right)}\)

Chúc bạn học tốt.

B1:

\(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)

Xét hiệu:

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\)

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

\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)

=> BĐT luôn đúng

*

Ta có:

\(a< b+c\Rightarrow a^2< ab+ac\)

\(b< a+c\Rightarrow b^2< ab+ac\)

\(c< a+b\Rightarrow a^2< ac+bc\)

Cộng từng vế bất đẳng thức ta được:

\(a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)

Vậy: \(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)

B2:

Ta có: \(a+b>c\) ; \(b+c>a\); \(a+c>b\)

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

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

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

Suy ra:

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

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

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

=> ĐPCM

d) \(\dfrac{a^2+b^2}{2}\) \(\ge\) \(\left(\dfrac{a+b}{2}\right)^2\)

<=> \(\dfrac{a^2+b^2}{2}\) \(\ge\) \(\dfrac{a^2+2ab+b^2}{4}\)

<=> 4(a² + b² ) \(\ge\) 2 ( a² + 2ab + b² )

<=> 4a² + 4b² \(\ge\) 2a² + 4ab +2b²

<=> 4a² + 4b² - 2a² - 4ab - 2b² \(\ge\) 0

<=> 2a² - 4ab + 2b² \(\ge\) 0

<=> a² -2ab +b² \(\ge\) 0

<=> (a-b)² \(\ge\) 0 ( luôn đúng)

=> \(\dfrac{a^2+b^2}{2}\) \(\ge\) \(\left(\dfrac{a+b}{2}\right)^2\)

Và dấu bằng xảy ra <=> a = b

e) Làm tương tự nhé! Có gì ko hiểu thì hỏi lại mk! Ok??

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

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

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

\(=\dfrac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}\)

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

\(=\dfrac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(a-c+b\right)}=\dfrac{a-c}{b-c}\left(đpcm\right)\)