Đặt: \(M=\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}=\Sigma_{cyc}\frac{a}{a^2+ab+bc+ca}\)

\(\Rightarrow M.\left(a+b+c\right)=3-\Sigma_{cyc}\frac{bc}{a^2+ab+bc+ca}\)

Đến đây t cần chứng minh:

 \(\frac{bc}{a^2+ab+bc+ca}+\frac{ca}{b^2+ab+bc+ca}+\frac{ab}{c^2+ab+bc+ca}\ge\frac{3}{4}\) (*)

Từ điều kiện ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\left(x,y,z>0\right)\)

\(\Rightarrow x+y+z=1\)

(*) \(\Leftrightarrow\frac{x^2}{\left(x+y\right)\left(z+x\right)}+\frac{y^2}{\left(x+y\right)\left(y+z\right)}+\frac{z^2}{\left(y+z\right)\left(z+x\right)}\ge\frac{3}{4}\)

Theo Cô-si: \(\frac{x^2}{\left(x+y\right)\left(z+x\right)}+\frac{9}{16}\left(x+y\right)\left(z+x\right)\ge\frac{3}{2}x\)

Nhứng phần kia tương tự

\(\Rightarrow\Sigma_{cyc}\frac{x^2}{\left(x+y\right)\left(z+x\right)}\ge\frac{3}{2}\left(x+y+z\right)-\frac{9}{16}\left[\left(x+y+z\right)^2+\left(xy+yz+zx\right)\right]\ge\frac{3}{4}\)

Lần trước làm không đúng hy vọng bây giờ gỡ lại được

nub

Bạn suy ra dòng 8 mk chưa hiểu, giải kĩ cho mk đc ko

\(\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca}\)

=\(\frac{1}{abc}.\left(\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca}\right)\)

=\(\frac{1}{a^5c+b^5c+abc}+\frac{1}{b^5a+c^5a+abc}+\frac{1}{c^5b+a^5b+abc}\)

\(\le\)\(\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}\)

Ta có : a³+b³=(a+b)(a²-ab+b²)\(\ge\)ab(a+b) (cosi)

Tương tự ta được:

b³+c³\(\ge bc\left(b+c\right)\)

c³+a³\(\ge ca\left(c+a\right)\)

Như vậy \(\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca}\)

\(\le\)\(\frac{1}{ab\left(a+b\right)+abc}+\frac{1}{bc\left(b+c\right)+abc}+\frac{1}{ca\left(c+a\right)+abc}\)

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

=\(\frac{1}{a+b+c}.\left(\frac{a+b+c}{ab+bc+ca}\right)=\frac{1}{ab+bc+ca}\le1\)

​mình tò mò muốn biết BĐT trên đẳng thức khi nào nhỉ

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

Từ \(abc=1\Rightarrow a=\frac{1}{bc}\) thay vào ta có:

\(A=\frac{\frac{1}{bc}}{\frac{1}{bc}\cdot b+\frac{1}{bc}+1}+\frac{b}{bc+b+1}+\frac{c}{c\cdot\frac{1}{bc}+c+1}\)

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

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

\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)

\(=\frac{1+b+bc}{bc+b+1}=1\)

Theo giả thiết, ta có: \(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\ge1\)\(\Leftrightarrow1-\frac{1}{a+b+1}+1-\frac{1}{b+c+1}+1-\frac{1}{c+a+1}\le2\)\(\Leftrightarrow\frac{a+b}{a+b+1}+\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\le2\)

Áp dụng bất đẳng thức Bunyakovsky dạng phân thức, ta được: \(\frac{a+b}{a+b+1}+\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\)\(=\frac{\left(a+b\right)^2}{\left(a+b\right)\left(a+b+1\right)}+\frac{\left(b+c\right)^2}{\left(b+c\right)\left(b+c+1\right)}+\frac{\left(c+a\right)^2}{\left(c+a\right)\left(c+a+1\right)}\)\(\ge\frac{\left(a+b+b+c+c+a\right)^2}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2+2\left(a+b+c\right)}\)

Từ đó suy ra \(\frac{\left(a+b+b+c+c+a\right)^2}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2+2\left(a+b+c\right)}\le2\)               \(\Leftrightarrow\left(a+b+b+c+c+a\right)^2\)                     \(\le2\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2+2\left(a+b+c\right)\right]\)

\(\Leftrightarrow a+b+c\ge ab+bc+ca\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi a = b = c = 1

Theo bđt Cauchy - Schwart ta có:

\(\text{Σ}cyc\frac{c}{a^2\left(bc+1\right)}=\text{Σ}cyc\frac{\frac{1}{a^2}}{b+\frac{1}{c}}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c}\)\(=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}\)

\(=\frac{\left(ab+bc+ca\right)^2}{abc\left(ab+bc+ca\right)+3a^2b^2c^2}\)

Đặt \(ab+bc+ca=x;abc=y\).

Ta có: \(\frac{x^2}{xy+3y^2}\ge\frac{9}{x\left(1+y\right)}\Leftrightarrow x^3+x^3y\ge9xy+27y^2\)

\(\Leftrightarrow x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\) ( luôn đúng )

Vậy BĐT đc CM. Dấu '=' xảy ra <=> a=b=c=1

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