Thay abc = 2 vào biểu thức A ta được:

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

Hàng thứ 2 phải sửa lại như vậy:

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

Bài 2:

a) \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)

\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)

\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)

\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)

Vì \(a+b+c=0\)

Nên a + b = -c (1)

Thay (1) vào A, ta được:

\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)

\(A=\dfrac{1}{abc}.3abc\)

\(A=3\)

b) \(B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)

\(B=\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(c^2+a^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\)

Vì \(a+b+c=0\)

Nên b + c = -a

=> ( b + c )² = (-a)²

=> b² + c² + 2bc = a²

=> b² + c² = a² - 2bc (1)

Tương tự ta có: c² + a² = b² - 2ac (2)

a² + b² = c - 2ab (3)

Thay (1), (2) và (3) vào B, ta được:

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

\(B=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ac}+\dfrac{c^2}{c^2-c^2+2ab}\)

\(B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)

\(B=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)

\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)

Mà \(a^3+b^3+c^3=3abc\) ( câu a )

\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)

\(\Rightarrow B=\dfrac{3}{2}\)

Bài 1:

a) GT: abc = 2

\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)

\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)

\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)

\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)

\(M=\dfrac{1+b+bc}{bc+b+1}\)

\(M=1\)

b) GT: abc = 1

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

\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)

\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)

\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)

\(N=\dfrac{1+b+bc}{bc+b+1}\)

\(N=1\)

\(\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(=\dfrac{a}{ab+a+2}+\dfrac{ab}{abc+ab+a}+\dfrac{2c}{ac+2c+abc}\)

\(=\dfrac{a}{ab+a+2}+\dfrac{ab}{2+ab+a}+\dfrac{2}{a+2+ab}\)

\(=\dfrac{ab+a+2}{ab+a+2}=1\)

ta có : \(A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(=\dfrac{a}{abc+ab+a}+\dfrac{b}{bc+b+1}+\dfrac{abc^2}{abc^2+abc+ac}\)

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

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

= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{abc+ab+a}\)+ \(\frac{2abc}{a^2bc+2abc+2ab}\)

= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{ab+a+2}\)+ \(\frac{2}{ab+a+2}\)   (vì  abc = 2  )

= \(\frac{ab+a+2}{ab+a+2}\)= 1

tại sao lại nhân vs a và ab z bn

nhân cả vế với abc ta có điều cần chứng minh

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

VT\(\ge\)\(\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{bc+ac+ab}{2}\)

=>(đpcm)

mấu chốt nằm ở đoạn chứng minh\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\) 

chỉ cần chứng minh được \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)sau đó áp dụng để chứng minh cái kia thôi cái này bạn thử tự chứng minh nhé

\(a,A=\dfrac{-3\left(2n-3\right)-8}{2n-3}=-3-\dfrac{8}{2n-3}\in Z\\ \Leftrightarrow2n-3\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow n\in\left\{1;2\right\}\left(n\in Z\right)\)

\(b,\dfrac{ab}{a+2b}=\dfrac{3}{2}\Leftrightarrow\dfrac{a+2b}{ab}=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{b}+\dfrac{2}{a}=\dfrac{2}{3}\\ \dfrac{bc}{b+2c}=\dfrac{4}{3}\Leftrightarrow\dfrac{b+2c}{bc}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{c}+\dfrac{2}{b}=\dfrac{3}{4}\\ \dfrac{ca}{c+2a}=3\Leftrightarrow\dfrac{c+2a}{ca}=\dfrac{1}{3}\Leftrightarrow\dfrac{1}{a}+\dfrac{2}{c}=\dfrac{1}{3}\)

Cộng vế theo vế \(\Leftrightarrow\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}=\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{7}{4}\)

\(\Leftrightarrow3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{7}{4}\\ \Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{7}{12}\\ \Leftrightarrow\dfrac{ab+bc+ca}{abc}=\dfrac{7}{12}\\ \Leftrightarrow T=\dfrac{12}{7}\)

Ta có \(ab+bc+ca=3abc\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\) thì ta có \(x,y,z>0;x+y+z=3\) và 

\(\sqrt{\dfrac{a}{3b^2c^2+abc}}=\sqrt{\dfrac{\dfrac{1}{x}}{3.\dfrac{1}{y^2z^2}+\dfrac{1}{xyz}}}=\sqrt{\dfrac{\dfrac{1}{x}}{\dfrac{3x+yz}{xy^2z^2}}}=\sqrt{\dfrac{y^2z^2}{3x+yz}}\) \(=\dfrac{yz}{\sqrt{3x+yz}}\) \(=\dfrac{yz}{\sqrt{x\left(x+y+z\right)+yz}}\) \(=\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}\)

Do đó \(T=\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)

Lại có \(\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\dfrac{yz}{2\left(x+y\right)}+\dfrac{yz}{2\left(x+z\right)}\)

Lập 2 BĐT tương tự rồi cộng theo vế, ta được \(T\le\dfrac{yz}{2\left(x+y\right)}+\dfrac{yz}{2\left(x+z\right)}+\dfrac{zx}{2\left(y+z\right)}+\dfrac{zx}{2\left(y+x\right)}\) \(+\dfrac{xy}{2\left(z+x\right)}+\dfrac{xy}{2\left(z+y\right)}\)

\(T\le\dfrac{yz+zx}{2\left(x+y\right)}+\dfrac{xy+zx}{2\left(y+z\right)}+\dfrac{xy+yz}{2\left(z+x\right)}\)

\(T\le\dfrac{x+y+z}{2}\) (do \(x+y+z=3\))

\(T\le\dfrac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\) \(\Leftrightarrow a=b=c=1\)

Vậy \(maxT=\dfrac{3}{2}\), xảy ra khi \(a=b=c=1\)

 (Mình muốn gửi lời cảm ơn tới bạn Nguyễn Đức Trí vì ý tưởng của bài này chính là bài mình vừa hỏi lúc nãy trên diễn đàn. Cảm ơn bạn Trí rất nhiều vì đã giúp mình có được lời giải này.)

 Bạn Lê Song Phương xem lại dùm nhé, thanks!

\(...\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\)

\(...\Rightarrow T\le2.3=6\)

\(\Rightarrow GTLN\left(T\right)=6\left(tạia=b=c=1\right)\)

\(A=\dfrac{x-4+5}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5}{\sqrt{x}-2}=\sqrt{x}+2+\dfrac{5}{\sqrt{x}-2}\)

\(=\sqrt{x}-2+\dfrac{5}{\sqrt{x}-2}+4\ge2\sqrt{\dfrac{5\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+4=4+2\sqrt{5}\)

\(A_{min}=4+2\sqrt{5}\) khi \(9+4\sqrt{5}\)

b.

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{l}{z}\right)\Rightarrow xyz=1\)

\(B=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)

\(B_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\Rightarrow a=b=c=1\)

khi 9+4\(\sqrt{5}\) là từ đâu ạ