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

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

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

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

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

1/ Đặt

\(\frac{a}{b^2}=x,\frac{b}{c^2}=y,\frac{c}{a^2}=z,xyz=1\)thì ta có

\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Leftrightarrow xy+yz+zx=x+y+z\)

\(\Leftrightarrow xyz-xy-yz-zx+x+y+z-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)

\(\Leftrightarrow x=1;y=1;z=1\)

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

\(\Leftrightarrow a=b^2;b=c^2;c=a^2\)

2/ Đặt

\(ab=x,bc=y,ca=z\) cần tính

\(P=\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\left(1+\frac{y}{x}\right)\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)

Xét \(x+y+z=0\)

\(\Rightarrow P=\frac{x+y}{x}.\frac{y+z}{y}.\frac{z+x}{z}=\frac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-1\)

Xét \(x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow x=y=z\)

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

mà nè đề bài câu 1 là abc=1 chứ có phải xyz=1 đâu

theo đề ra ta có \(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

                          \(\Leftrightarrow2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=0\)

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

ta có đề <=>\(\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(a+c\right)=3abc\)

mà a+b+c=0=>a+b=-c,b+c=-a,a+c=-b thay vào biểu thức trên

             \(\Leftrightarrow-3\left(-a\right)\left(-b\right)\left(-c\right)=3abc\)

              <=> \(3abc=3abc\)(hiển nhiên đúng)

vậy BĐT được chứng minh

đúng thì đúng nhưng cần sửa

\(2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=0\)

<=>\(\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}=0\)

<=>\(\frac{a+b+c}{abc}=0\)

do a,b,c khác 0 nên abc khác 0

=> a+b+c=0

=> a+b= -c

<=> \(\left(a+b\right)^3=\left(-c\right)^3\)

<=>\(a^3+b^3+3ab\left(a+b\right)=-c^3\)

<=>   \(a^3+b^3-3abc=-c^3\)(do ab = -c)

<=> \(a^3+b^3+c^3=3abc\)(đpcm)

bạn nguyên x thị lan hương trình bày còn kém

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

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

\(\Leftrightarrow2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=0\)

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

\(\Leftrightarrow\frac{abc}{ab}+\frac{abc}{bc}+\frac{abc}{ac}=0\left(\text{​nhân cả hai về với abc}\right)\)

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

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

\(\Leftrightarrow\left(a+b\right)^3=-c^3\)

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3a^2b-3ab^2\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(-c\right)\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

Vậy đpcm

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 \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)

Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)

Ta có:

\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)

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

Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)

\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)

Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)

Ta có:\(a+b+c=0\)

\(\Rightarrow\left(a+b\right)^3=-c^3\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)

Mách mk nốt 2 bài kia vs

chiju