Dat \(\hept{\begin{cases}A=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\\B=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\end{cases}}\)

Ta co:\(A=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge2+2+2=6\left(1\right)\)

\(B=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}-3\)

\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\ge\left(a+b+c\right).\frac{9}{2\left(a+b+c\right)}-3=\frac{3}{2}\left(2\right)\)

Cong ve voi ve cua (1) va (2) ta duoc:

\(P=A+B\ge6+\frac{3}{2}=\frac{15}{2}\)

Dau '=' xay ra khi \(a=b=c\)

Chứng minh ĐBT:\(\frac{b}{a}+\frac{a}{b}\ge2\left(a,b\ne0\right)\)(Dấu "="\(\Leftrightarrow a=b=1\))

Ta có: \(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow\frac{a^2+b^2}{ab}\ge2\)

\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}\ge2\left(đpcm\right)\)

Vậy \(\frac{b+c}{a}+\frac{a}{b+c}\ge2\)

\(\frac{a+c}{b}+\frac{b}{c+a}\ge2\)

\(\frac{a+b}{c}+\frac{c}{b+a}\ge2\)

\(\Rightarrow P\ge6\)

Vậy \(P_{min}=6\Leftrightarrow\hept{\begin{cases}a=b+c\\b=a+c\\c=a+b\end{cases}}\)

dấu = ko xảy ra => tất cả sai:)

Đầu tiền dùng AM-GM cm tổng 3 phân thức đầu >= 6

 tổng 3 phân thức còn lại >= 3/2(bđt nesbit) .vậy là xong

vì a,b,c dương => a+b khác 0 

                             b+c khác 0 

                              a+c khác 0 

áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

\(=\frac{1}{2}\)

vậy E = \(\frac{1}{2}\)

Áp dụng BĐT AM-GM ta có: \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)

Tương tự ta có: \(\frac{b^2}{c+a}+\frac{c+a}{4}\ge b;\) \(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)

Cộng 3 BĐT trên theo vế thì được: 

\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+\frac{a+b+c}{2}\ge a+b+c\)

\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c\ge\frac{3\left(a+b+c\right)}{2}\)

\(\Leftrightarrow\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}\ge\frac{3\left(a+b+c\right)}{2}\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)\(\Rightarrow E\ge\frac{3}{2}\).

Vậy \(Min\) \(E=\frac{3}{2}\). Đẳng thức xảy ra <=> a=b=c. 

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

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

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

Xảy ra khi \(a=b=c=\frac{1}{3}\)

Áp dụng BĐT Svac - xơ:

\(T=\frac{a}{a^2+8bc}+\frac{b}{b^2+8ca}+\frac{c}{c^2+8ab}\)

\(=\frac{a^2}{a^3+8abc}+\frac{b^2}{b^3+8abc}+\frac{c^2}{c^3+8abc}\)\(\ge\frac{\left(a+b+c\right)^2}{a^3+b^3+c^3+24abc}\)

Ta lại có: \(\left(a+b+c\right)^3=a^3+b^3+c^3+\)\(3\left(a+b+c\right)\left(ab+bc+ca\right)-3abc\)

\(\ge a^3+b^3+c^3+27\sqrt[3]{abc}.\sqrt[3]{\left(abc\right)^2}-3abc=\)\(a^3+b^3+c^3+24abc\)

Lúc đó: \(T\ge\frac{1}{a+b+c}=1\)

(Dấu "="\(\Leftrightarrow a=b=c=\frac{1}{3}\))

Cho tớ sửa đề 

tử của ba cái là mũ 2 lên hết nha

\(T=\frac{a^2}{a^2+8bc}+\frac{b^2}{b^2+8ca}+\frac{c^2}{c^2+8ab}\)

\(\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac+6\left(ab+bc+ac\right)}\)

\(\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+6.\frac{\left(a+b+c\right)}{3}^2}\)

\(=\frac{1}{1+\frac{6}{3}}=\frac{1}{3}\)

Dấu "=" xảy ra <=> a = b = c = 1/3 

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

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

Ta đặt \(\frac{b}{a}=x;\frac{c}{b}=y\Rightarrow\frac{c}{a}=xy\)

\(P=\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}+\frac{xy}{4}\)

Lại có \(\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}\ge\frac{1}{xy+1}\)

Thật vậy, bđt trên tương đương với:

 \(\left(xy+1\right)\left[\left(1+x\right)^2+\left(1+y\right)^2\right]\ge\left(1+x\right)^2\left(1+y\right)^2\)

\(\Leftrightarrow\left(xy+1\right)\left(x^2+y^2+2x+2y+2\right)\ge\left(x^2+2x+1\right)\left(y^2+2y+1\right)\)

\(\Leftrightarrow x^2y+y^2x-x^2y^2-2xy+1\ge0\)

\(\Leftrightarrow xy\left(x-y\right)^2+\left(xy-1\right)^2\ge0\)luôn đúng

Suy ra: \(P\ge\frac{1}{xy+1}+\frac{xy}{4}=\frac{1}{xy+1}+\frac{xy+1}{4}-\frac{1}{4}\) 

           \(P\ge2\sqrt{\frac{1}{xy+1}\frac{xy+1}{4}}-\frac{1}{4}\left(AM-GM\right)\)   

                \(=1-\frac{1}{4}=\frac{3}{4}\)

Đẳng thức xảy ra khi a=b=c=1