Ta có : a + b + c = abc

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

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

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

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

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=4\)  

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\) 

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\) 

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

\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{a+b+c}{abc}=\frac{abc}{abc}=1\left(\text{vì }a+b+c=abc\right)\)

\(\text{Lại có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.1=2\left(\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\right)\)

Vậy ... 

đây là toán 8 thật sao ạ!

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac. 
(1/a + 1/b + 1/c)² = 1/a² + 1/b² + 1/c² + 2(1/ab + 1/bc + 1/ac) = 4 
<=> 1/a² + 1/b² + 1/c² + 2(bcac + abac + abbc)/(a²b²c²) = 4 
<=> 1/a² + 1/b² + 1/c² + 2abc(a + b + c)/(a²b²c²) = 4 
<=> 1/a² + 1/b² + 1/c² + 2 = 4 
(vi` abc(a + b + c) = a² b² c²) 
<=> 1/a² + 1/b² + 1/c² = 2 !!

Cho abc=0 thì không chứng minh được, a+b+c=0 là đủ rồi

Ta có: a+b+c=0 => a+b=-c

=>(a+b)²=(-c)²

=>a²+2ab+b²=c²

=>a²+b²-c²=-2ab

Tương tự ta có: b²+c²-a²=-2bc ; c²+a²-b²=-2ca

=>\(\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}=-\frac{1}{2bc}-\frac{1}{2ca}-\frac{1}{2ab}=\frac{a+b+c}{-2abc}=0\) (đpcm)

Cho \(abc=0\)thì không chứng minh được, \(a+b+c=0\)là đủ rồi.

Ta có: \(a+b+c=0\Rightarrow a+b=-c\)

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

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

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

Tương tự ta có: \(b^2+c^2-a^2=-2ab;c^2+a^2-b^2=-2ca\)

\(\Rightarrow\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}=-\frac{1}{2bc}-\frac{1}{2ca}-\frac{1}{2ab}=\frac{a+b+c}{-2abc}=0\)

\(\Leftrightarrow\dfrac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\le1\)

\(\Leftrightarrow\dfrac{ab+bc+ca+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\le1\)

\(\Leftrightarrow ab+bc+ca+12\le2\left(ab+bc+ca\right)+9\)

\(\Leftrightarrow ab+bc+ca\ge3\)

Hiển nhiên đúng do: \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}=3\)

Vì abc=1 , ta đặt \(a=\dfrac{x}{y};b=\dfrac{y}{z};c=\dfrac{z}{x}\)

Điều phải chứng minh tương đương với:

\(\dfrac{1}{2+\dfrac{x}{y}}+\dfrac{1}{2+\dfrac{y}{z}}+\dfrac{1}{2+\dfrac{z}{x}}\le1\\ \Leftrightarrow\dfrac{y}{2y+x}+\dfrac{z}{2z+y}+\dfrac{x}{2x+z}\le1\\ \Leftrightarrow\dfrac{2y}{2y+x}+\dfrac{2z}{2z+y}+\dfrac{2x}{2x+z}\le2\\ \Leftrightarrow\dfrac{x}{2y+x}+\dfrac{y}{2z+y}+\dfrac{z}{2x+z}\ge1\left(1\right)\)

Áp dụng bất đẳng thức bunhiacopxki dạng phân thức ta có:

\(\dfrac{x}{2y+x}+\dfrac{y}{2z+x}+\dfrac{z}{2x+z}=\dfrac{x^2}{x^2+2xy}+\dfrac{y^2}{y^2+2zx}+\dfrac{z^2}{z^2+2xy}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)

=> bài toán được chứng minh

Dấu bằng xảy ra khi x=y=z=1 <=>a=b=c=1

Thôi câu đó mình làm được rồi, các bạn giúp mình câu này nha

Cho \(a>b\ge0\). CMR: \(\dfrac{a^4+b^4}{a^4-b^4}-\dfrac{ab}{a^2-b^2}+\dfrac{a+b}{2\left(a-b\right)}\ge\dfrac{3}{2}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\\ \to ab+bc+ca=abc=1\)

Ta có \(A=\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)\)

\(\to A=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)

\(\to A=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

Vì $a,b,c\in \mathbb{Q}\to A\in \mathbb{Q}$

Bài toán cơ bản:

\(abc=1\Rightarrow\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\) 

Bunhiacopxki:

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

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

Dấu "=" xảy ra khi \(a=b=c\)

Cách 2:

Do \(abc=1\), đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)

Ta có \(\dfrac{a}{\left(ab+a+1\right)^2}=\dfrac{\dfrac{x}{y}}{\left(\dfrac{x}{z}+\dfrac{x}{y}+1\right)^2}=\dfrac{\dfrac{x}{y}.y^2z^2}{\left(xy+yz+zx\right)^2}=\dfrac{xyz^2}{\left(xy+yz+zx\right)^2}\)...

Từ đó, BĐT cần chứng minh trở thành:

\(\dfrac{xyz^2}{\left(xy+yz+zx\right)^2}+\dfrac{x^2yz}{\left(xy+yz+zx\right)^2}+\dfrac{xy^2z}{\left(xy+yz+zx\right)^2}\ge\dfrac{1}{\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}}\)

\(\Leftrightarrow xyz\left(x+y+z\right)\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\ge\left(xy+yz+zx\right)^2\)

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

Thật vậy, áp dụng BĐT Bunhiacopxki:

\(\left(z+x+y\right)\left(x^2z+y^2x+z^2y\right)\ge\left(\sqrt{zx^2z}+\sqrt{xy^2x}+\sqrt{yz^2y}\right)^2=\left(xy+yz+zx\right)^2\) (đpcm)