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 )

Em thử nha, có gì sai bỏ qua ạ.

Đề cho gọn,Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\) thì \(xy+yz+zx=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=0\) 

Và \(x+y+z=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}=0\)

Ta có: \(VT=\sqrt{x^2+y^2+z^2}=\sqrt{\left(x+y+z\right)^2-2\left(xy+yz+zx\right)}=0\) (1)

Mặt khác,ta có \(VT=\left|x+y+z\right|=0\) (2)

Từ (1) và (2) ta có đpcm

​Dòng cuối phải là

VP=|x+y+z|=0 

đúng không????

uk,thì e viết nhầm -_-"

Áp dụng BĐT Cosi ta có \(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab}\ge2\sqrt{\frac{ab}{a^2+b^2}.\frac{a^2+b^2}{4ab}}=1\)

Tương tự \(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{4bc}\ge1\) \(\frac{ca}{c^2+a^2}+\frac{c^2+a^2}{4ca}\ge1\)

Khi đó BĐT sẽ được chứng minh nếu ta chỉ ra được

\(\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\left(\frac{a^2+b^2}{4ab}+\frac{b^2+c^2}{4bc}+\frac{c^2+a^2}{4ca}\right)\ge\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\left(\frac{a}{4b}+\frac{b}{4a}+\frac{b}{4c}+\frac{c}{4b}+\frac{a}{4c}+\frac{c}{4a}\right)\right)\ge\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}\left(\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}-\frac{a+c}{b}-\frac{b+c}{a}-\frac{c+a}{b}\right)\ge\frac{3}{4}\)(do \(a+b+c=1\))

\(\Leftrightarrow\frac{3}{4}\ge\frac{3}{4}\) luôn đúng. Từ đó suy ba BĐT được chứng minh. Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

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

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

\(=\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac{2}{abc}.\left(a+b+c\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)(Vì a + b + c = 0)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\) (đpcm)

Áp dụng BĐT Cauchy ta có : \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\Rightarrow\frac{1}{\frac{1}{a}+\frac{1}{b}}\le\frac{\sqrt{ab}}{2}\)

Thiết lập tương tự và thu lại ta có :

\(\Rightarrow VP\le4\left(\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2}\right)=2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(1\right)\)

Áp dụng BĐT Cauchy ta có : \(a+b\ge2\sqrt{ab}\)
\(\Rightarrow\left(a+b+\frac{1}{2}\right)^2\ge\left(2\sqrt{ab}+\frac{1}{2}\right)^2\ge2.2\sqrt{ab}.\frac{1}{2}=2\sqrt{ab}\)

Thiết lập tương tự và thu lại ta có ;

\(\Rightarrow VT\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(2\right)\)

Từ (1) và (2)  suy ra

\(VT\ge VP\)

\(\Rightarrowđpcm\)

Chúc bạn học tốt !!!

Áp dụng bđt Cauchy ta có : \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\Rightarrow\frac{1}{\frac{1}{a}+\frac{1}{b}}\le\frac{\sqrt{ab}}{2}\)

Thiết lập tương tự và thu lại ta có :

\(\Rightarrow VP\le4\left(\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2}\right)=2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(1\right)\)

Áp dụng bđt Cauchy ta cso :
\(a+b\ge2\sqrt{ab}\)

\(\Rightarrow\left(a+b+\frac{1}{2}\right)^2\ge\left(2\sqrt{ab}+\frac{1}{2}\right)^2\ge2.2\sqrt{ab}.\frac{1}{2}=2\sqrt{ab}\)

Thiết lập tương tự và thu lại ta có :

\(\Rightarrow VT\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(2\right)\)

Từ (1) và (2) 

\(VT\ge VP\)

\(\Rightarrowđpcm\)

Chúc bạn học tốt !!!

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

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

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)

\(\Leftrightarrow a+b+c=0\)

Xét : \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right).\left(b+c\right).\left(c+a\right)=-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\) luôn chia hết cho 3

khó quá ta

Đặt : x/a = m ; y/b = n ; z/c = p

=> m+n+p = 1 ; 1/m+1/n+1/p=0

1/m+1/n+1/p=0

<=> mn+np+pm/mnp=0

<=> mn+np+pm=0

<=> 2mn+2np+2pm=0

Xét : 1 = (m+n+p)^2 = m^2+n^2+p^2+2mn+2np+2pm = m^2+n^2+p^2

=> x^2/a^2+y^2/b^2+z^2/c^2 = 1

=> ĐPCM

Tk mk nha

Ta có: \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Rightarrow\frac{ayz}{xyz}+\frac{bxz}{xyz}+\frac{cxy}{xyz}=0\)

\(\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\)

\(\Rightarrow ayz+bxz+cxy=0\left(1\right)\)

Mặt khác: \(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+ayz+bxz}{abc}\right)=1\left(2\right)\)

Thay (1) vào (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{0}{abc}=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)