Từ \(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}\Rightarrow\left(10a+b\right).\left(b+c\right)=\left(10b+c\right).\left(a+b\right)\)

\(\Rightarrow10ab+b^2+10ac+bc=10ab+ac+10b^2+bc\)

\(\Rightarrow b^2+10ac=ac+10b^2\)

\(\Rightarrow10ac-ac=10b^2-b^2\)

\(\Rightarrow9ac=9b^2\)

\(\Rightarrow ac=b^2\Rightarrow\frac{a}{b}=\frac{b}{c}\left(đpcm\right)\)

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

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

<=> \(\frac{a.10+b}{b.10+c}=\frac{a+b}{b+c}\)

Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{a.10+b}{b.10+c}=\frac{a+b}{b+c}=\frac{\left(10a+b\right)-\left(a+b\right)}{\left(10b+c\right)-\left(b+c\right)}=\frac{9a}{9b}=\frac{a}{b}\)

Áp dụng dãy tỉ số bằng nhau ta có:

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

=> \(\frac{a}{b}=\frac{b}{c}\)

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

sau đó chứng minh tương tự và cộng theo từng vế thôi 

Ta có

\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)

\(\Rightarrow\left\{\begin{matrix}\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}\\\frac{b}{c-a}=-\frac{a}{b-c}-\frac{c}{a-b}\\\frac{c}{a-b}=-\frac{a}{b-c}-\frac{b}{c-a}\end{matrix}\right.\) (1)

Mà

\(\left\{\begin{matrix}\frac{a}{\left(b-c\right)^2}=\frac{a}{b-c}.\frac{1}{b-c}\\\frac{b}{\left(c-a\right)^2}=\frac{b}{c-a}.\frac{1}{c-a}\\\frac{c}{\left(a-b\right)^2}=\frac{c}{a-b}.\frac{1}{a-b}\end{matrix}\right.\)

Ta có : \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)

\(\Rightarrow\frac{a}{b-c}.\frac{1}{b-c}+\frac{b}{c-a}.\frac{1}{c-a}+\frac{c}{a-b}.\frac{1}{a-b}=0\)

Thay điều (1) vào biểu thức ta có :

\(\frac{a}{b-c}.\frac{1}{b-c}+\frac{b}{c-a}.\frac{1}{c-a}+\frac{c}{a-b}.\frac{1}{a-b}=0\)

\(\Rightarrow\left(-\frac{b}{c-a}-\frac{c}{a-b}\right).\frac{1}{b-c}+\left(-\frac{a}{b-c}-\frac{c}{a-b}\right).\frac{1}{c-a}+\left(-\frac{a}{b-c}-\frac{b}{c-a}\right).\frac{1}{a-b}=0\)

\(\Rightarrow-\frac{b}{\left(c-a\right)\left(b-c\right)}-\frac{c}{\left(a-b\right)\left(b-c\right)}-\frac{a}{\left(b-c\right)\left(c-a\right)}-\frac{c}{\left(a-b\right)\left(c-a\right)}-\frac{a}{\left(b-c\right)\left(a-b\right)}-\frac{b}{\left(c-a\right)\left(a-b\right)}=0\)

\(\Rightarrow-\frac{b}{\left(c-a\right)\left(b-c\right)}-\frac{a}{\left(c-a\right)\left(b-c\right)}-\frac{c}{\left(a-b\right)\left(b-c\right)}-\frac{a}{\left(a-b\right)\left(b-c\right)}-\frac{c}{\left(c-a\right)\left(a-b\right)}-\frac{b}{\left(c-a\right)\left(a-b\right)}=0\)

\(\Rightarrow-\frac{b-a}{\left(c-a\right)\left(b-c\right)}-\frac{c-a}{\left(a-b\right)\left(b-c\right)}-\frac{c-b}{\left(c-a\right)\left(a-b\right)}=0\)

\(\Rightarrow-\left[\frac{b+a}{\left(c-a\right)\left(b-c\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}+\frac{c+b}{\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow-\left[\frac{\left(b+a\right)\left(a-b\right)^2\left(b-c\right)\left(c-a\right)+\left(c+a\right)\left(c-a\right)^2\left(b-c\right)\left(a-b\right)+\left(c+b\right)\left(b-c\right)^2\left(c-a\right)\left(a-b\right)}{\left(b-c\right)^2\left(c-a\right)^2\left(a-b\right)^2}\right]=0\)

\(\Rightarrow-\left\{\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left[\left(b+a\right)\left(a-b\right)+\left(c+a\right)\left(c-a\right)+\left(b+c\right)\left(b-c\right)\right]}{\left(b-c\right)^2\left(c-a\right)^2\left(a-b\right)^2}\right\}=0\)

\(\Rightarrow-\left[\frac{\left(b+a\right)\left(b-a\right)+\left(c+a\right)\left(c-a\right)+\left(b+c\right)\left(b-c\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow-\left[\frac{\left(a^2-b^2\right)+\left(c^2-a^2\right)+\left(b^2-c^2\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow-\left[\frac{\left(-b^2+b^2\right)+\left(-a^2+a^2\right)+\left(-c^2+c^2\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow-\left[\frac{0}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow0=0\) ( đpcm )

hình như là a²=bc mà bạn]

hình như đề sai b²=ac ko pải a²=ad đâu

3, \(\sqrt{\frac{a}{b+c}}=\sqrt{\frac{a^2}{a\left(b+c\right)}}\Rightarrow\frac{1}{\sqrt{\frac{a}{b+c}}}=\sqrt{\frac{a\left(b+c\right)}{a^2}}.\)

Áp dụng bất đẳng thức Cô si ta có : \(\sqrt{\frac{a\left(b+c\right)}{a^2}}\le\frac{a+b+c}{2a}\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\left(1\right).\)

Chứng minh tương tự ta có : \(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\left(2\right).\);  \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\left(3\right).\)

Cộng vế với vế các bất đẳng thức cùng chiều ta được: 

\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2.\)( đpcm )

dấu " = " xẩy ra khi a = b = c > 0

Đặt \(\frac{a}{b}=x;\frac{b}{c}=y;\frac{c}{a}=z\)

\(\Rightarrow xyz=\frac{a}{b}.\frac{b}{c}.\frac{c}{a}=1\)

Bất đẳng thức đã cho tương đương với: \(\Leftrightarrow x^2+y^2+z^2\ge\frac{z}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)

\(\Leftrightarrow2.\left(x^2+y^2+z^2\right)-2.\left(xy+yz+zx\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\left(\forall x;y;z\right)\)

Dấu "=" xảy ra khi \(\Leftrightarrow x=y=z\Rightarrow a=b=c\left(đpcm\right)\) 

b²=ac mà bạn

b) a²=ac\(\Rightarrow\) \(\frac{a}{b}=\frac{b}{c}\)

c²=bd\(\Rightarrow\) \(\frac{b}{c}=\frac{c}{d}\)

\(\Rightarrow\)\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) = \(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\) = \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

=\(\frac{a.b.c}{b.c.d}=\frac{a}{d}\)

=> đpcm

với x=y=z khác 0 và a,b,c khác nhau là 1 số bất kỳ khác 0 thì (1) thỏa mãn và (2) không thỏa mãn

=> Không thể CM

ta có: \(\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}\)

\(\Rightarrow\frac{a}{x^2-yz}=\frac{b}{y^2-zx}=\frac{c}{z^2-xy}\) (*)

\(\Rightarrow\frac{a^2}{\left(x^2-yz\right)^2}=\frac{bc}{\left(y^2-zx\right).\left(z^2-xy\right)}=\frac{a^2-bc}{\left(x^2-yz\right)^2-\left(y^2-zx\right).\left(z^2-xy\right)}\)

\(=\frac{a^2-bc}{x^4-3x^2yz+xy^3+xz^3}=\frac{a^2-bc}{x.\left(x^3-3xyz+y^3+z^3\right)}\)

\(\Rightarrow\frac{a^2-bc}{x}=\frac{a^2}{\left(x^2-yz\right)^2}.\left(x^3-3xyz+y^3+z^3\right)\)

Làm tương tự như trên. ta có:

\(\frac{b^2-ca}{y}=\frac{b^2}{\left(y^2-zx\right)^2}.\left(x^3-3xyz+y^3+z^3\right)\)

\(\frac{c^2-ab}{z}=\frac{c^2}{\left(z^2-xy\right)^2}.\left(x^3-3xyz+y^3+z^3\right)\)

Từ (*) \(\Rightarrow\frac{a^2-bc}{x}=\frac{b^2-ca}{y}=\frac{c^2-ab}{z}\left(đpcm\right)\)