Thay abc = 1 vào biểu thức ta có

\(\frac{a.abc}{ab+abc.a+abc}+\frac{b}{bc+b.acb+abc}+\frac{c}{ac+c+1}\)

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

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

= \(\frac{ac}{\left(ac+c+1\right)}+\frac{1}{\left(ac+c+1\right)}+\frac{c}{ac+c+1}\)

= \(\frac{ac+c+1}{ac+c+1}\)

= 1 (đpcm)

Nếu có gì không hiểu nhớ nt cho mình nha

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

\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{abc}{a\cdot abc+abc+ab}\)

\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{a+1+ab}\)

\(=\frac{ab+a+1}{ab+a+1}=1\)

Em xem phần trả lời của bạn Giang nhé Câu hỏi của Vu Hoang - Toán lớp 8 - Học toán với OnlineMath

1 cách khác:

Đặt \(a=\frac{x}{y};b=\frac{y}{z};c=\frac{z}{x}\Rightarrow abc=1\left(TMGT\right)\)

Khi đó:

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

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

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

\(=\frac{xz}{xy+yz+zx}+\frac{xy}{xy+yz+zx}+\frac{yz}{xy+yz+zx}=\frac{xy+yz+zx}{xy+yz+zx}=1\)

Ta chứng minh được

\(a^4+b^4\ge ab\left(a^2+b^2\right)\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

\(\Rightarrow P\le\sum\frac{ab}{ab\left(a^2+b^2\right)+ab}=\sum\frac{1}{a^2+b^2+1}\)

Đặt \(\left(a^2;b^2;c^2\right)=\left(x^3;y^3;z^3\right)\Rightarrow xyz=1\)

Ta lại chứng minh được:

\(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\)

\(\Rightarrow P\le\sum\frac{1}{x^3+y^3+1}\le\sum\frac{xyz}{xy\left(x+y\right)+xyz}=\sum\frac{z}{x+y+z}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Đây là bài thi vào 10 của Thanh Hóa thì phải

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

\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}=1\)

Thay \(abc=1\) vào biểu thức ta có :

 \(\frac{a.abc}{ab+abc.a+abc}+\frac{b}{bc+b.acb+abc}+\frac{c}{ac+c+1}\)

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

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

\(=\frac{ac}{\left(ac+c+1\right)}+\frac{1}{\left(ac+c+1\right)}+\frac{c}{ac+c+1}\)

\(=\frac{ac+c+1}{ac+c+1}\)

\(=1\left(đpcm\right)\)

Chúc bạn học tốt !!!

do abc=1 nên \(\frac{a}{ab+a+1}\)=\(\frac{a}{ab+a+abc}\)=\(\frac{a}{a\left(bc+b+1\right)}\)=\(\frac{1}{bc+b+1}\)

\(\frac{c}{ac+c+1}\)=\(\frac{bc}{abc+bc+b}\)(nhân cả 2 vế cho b)=\(\frac{bc}{bc+b+1}\)

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

\(~Angle\)\(Darkness~\)

Ta có: \(ab+bc+ca=abc\)

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

Đặt: \(A=\frac{a}{bc\left(a+1\right)}+\frac{b}{ca\left(b+1\right)}+\frac{c}{ab\left(c+1\right)}\)

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

Đặt: \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow x+y+z=1\)

\(A=\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\)

Ta có: \(\frac{xy}{z+1}=\frac{xy}{\left(z+x\right)+\left(z+y\right)}\le\frac{1}{4}\left(\frac{xy}{x+z}+\frac{xy}{y+z}\right)\)

Chứng minh tương tự ta được:

\(\frac{yz}{x+1}\le\frac{yz}{x+y}+\frac{yz}{x+z}\)

\(\frac{zx}{y+1}\le\frac{zx}{x+y}+\frac{zx}{y+z}\)

Cộng vế với vế:

\(\Rightarrow A\le\frac{1}{4}\left(x+y+z\right)=\frac{1}{4}\left(đpcm\right)\)

1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)

\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

Cho o dong 2 la x,y,z nhe,ghi nham

Đặt \(T=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ac}\) (*)

Ta có: \(abc=1\Rightarrow c=\frac{1}{ab}\).Thay vào (*) ta có:

\(T=\frac{1}{1+a+ab}+\frac{1}{1+b+\frac{1}{a}}+\frac{1}{1+\frac{1}{ab}+\frac{1}{b}}\)

\(=\frac{1}{1+a+ab}+\frac{1}{\frac{a+ab+1}{a}}+\frac{1}{\frac{ab+1+a}{ab}}\)

\(=\frac{1}{1+a+ab}+\frac{a}{a+ab+1}+\frac{ab}{ab+1+a}\)

\(=\frac{1+a+ab}{1+a+ab}=1=VP\) (Đpcm)