\(M=\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-a\right)}\)

Đánh giá đại diện: \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}-\frac{1}{a-c}\)

Tương tự: \(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}-\frac{1}{b-a}\)

                   \(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}-\frac{1}{c-b}\)

\(\Rightarrow M=\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}\)

\(\Rightarrow M=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\)

\(\Rightarrow M=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2N\left(đpcm\right)\)

Áp dụng BĐT Bunhiacopxki, ta có: 

\(\left(a+b+c\right)\left(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right)\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\)

Mà \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+1}=1\)

\(\Rightarrow\left(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right)\left(a+b+c\right)\ge1\) 

\(\Rightarrow\frac{a}{\left(ab+b+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)

\(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)

ta có  \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)

\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}=1\)

đặt \(H=\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\)

áp dụng bất đẳng thức bunhiacopxki  ta có 

\(H\left(a+b+c\right)\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\right)^2=1\)

\(\Rightarrow H\ge\frac{1}{a+b+c}\)

hay  \(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)

Xin lỗi lúc này do thày nhìn nhầm nên nghĩ câu 2 sai đề. Để đền bù thiệt hại, xin giải lại cả hai bài cho em

Cả hai bài toán này đều sử dụng bất đẳng thức Cauchy-Schwartz. Em xem link dưới đây để biết rõ hơn: http://olm.vn/hoi-dap/question/174274.html

Câu 1. Theo bất đẳng thức Cauchy-Schwartz ta có

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

\(\ge\frac{\left(1+1+1\right)^2}{2\left(a+b+c\right)+\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)}=\frac{9}{2\left(a+b+c\right)+\frac{a^2b^2+b^2c^2+c^2a^2}{abc}}=\frac{9abc}{2abc\left(a+b+c\right)+\left(a^2b^2+b^2c^2+c^2a^2\right)}\)

\(=\frac{9abc}{\left(ab+bc+ca\right)^2}=\frac{9abc}{9}=abc.\)

Vậy ta có điều phải chứng minh.

Câu 2.  Tiếp tục sử dụng bất đẳng thức Cauchy-Schwartz

\(\frac{8}{2a+b}=\frac{4}{a+\frac{b}{2}}\le\frac{1}{a}+\frac{1}{\frac{b}{2}}=\frac{1}{a}+\frac{2}{b}.\)

Tương tự, \(\frac{48}{3b+2c}=\frac{16}{b+\frac{2c}{3}}\le4\left(\frac{1}{b}+\frac{1}{\frac{2c}{3}}\right)=\frac{4}{b}+\frac{6}{c},\) và \(\frac{12}{c+3a}=\frac{4}{\frac{c}{3}+a}\le\frac{1}{\frac{c}{3}}+\frac{1}{a}=\frac{3}{c}+\frac{1}{a}.\)

Cộng ba bất đẳng thức lại ta được

\(\frac{8}{2a+b}+\frac{48}{3b+2c}+\frac{12}{c+3a}\le\left(\frac{1}{a}+\frac{2}{b}\right)+\left(\frac{4}{b}+\frac{6}{c}\right)+\left(\frac{3}{c}+\frac{1}{a}\right)=\frac{2}{a}+\frac{6}{b}+\frac{9}{c}.\)    (ĐPCM).

\(\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)abc=\frac{3}{4}8\Rightarrow\frac{abc}{a^2}+\frac{abc}{b^2}+\frac{abc}{c^2}=\frac{3.8}{4}\Leftrightarrow\)\(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}=6\)

lm đc trước rồi nhưng cũng k cho