cho 2 biểu thức mà c/m 1 biểu thức M là sao

Biểu thức N vứt sọt à hay làm cái j v :V

tớ cũng nghĩ vậy nhưng mãi sau mới biết chứng minh M =N rồi chứng minh N >=(a+b+c)/8 để suy ra M  >=(a+b+c)/8

Ta có : \(\frac{a^2-bc}{a}+\frac{b^2-ac}{b}+\frac{c^2-ab}{c}=0\)

=> \(a-\frac{bc}{a}+b-\frac{ac}{b}+c-\frac{ab}{c}=0\)

=> \(a+b+c=\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\)

=> \(a+b+c=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

=> \(\frac{a+b+c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

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

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

=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{bc}-\frac{2}{ac}-\frac{2}{ac}=0\)

=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{a^2}-\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{b^2}-\frac{1}{bc}+\frac{1}{c^2}\right)=0\)

=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{a}-\frac{1}{c}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2=0\)

=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{a}-\frac{1}{c}=0\\\frac{1}{b}-\frac{1}{c}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}\\\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{c}\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\left(\text{đpcm}\right)\)

xí câu 1:))

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge\frac{\left(x+y\right)^2}{x+y-2}\)(1)

Đặt a = x + y - 2 => a > 0 ( vì x,y > 1 )

Khi đó \(\left(1\right)=\frac{\left(a+2\right)^2}{a}=\frac{a^2+4a+4}{a}=\left(a+\frac{4}{a}\right)+4\ge2\sqrt{a\cdot\frac{4}{a}}+4=8\)( AM-GM )

Vậy ta có đpcm

Đẳng thức xảy ra <=> a=2 => x=y=2

duyhung723 nè

Ta có : \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)

\(\Rightarrow a^2c+b^2c-ab^2-ac^2=0\)

\(\Rightarrow a\left(ac-b^2\right)-c\left(ac-b^2\right)=0\)

\(\Rightarrow\left(a-c\right)\left(ac-b^2\right)=0\)

\(\Rightarrow ac=b^2\) ( do \(a\ne c\) )

\(\Rightarrow\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{c}{b}=\frac{b}{a}=\frac{a}{c}\)

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

tu gia thiet co dc ab+bc+ca=0.Dat ab=x,bc=y,ca=z. Can chung minh x^3+y^3+z^3=3xyz

ta có: a+b+c=0↔a+b=-c

↔a³+3a²b+3ab²+b³=-c³

↔a³+b³+c³=-3ab(a+b)=-3ab.(-c)=3abc

→A=\(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)