giả sử a\(\le\)b \(\le\)c.

khi đó \(\frac{a}{b+c}\le\frac{b}{c+a}\le\frac{c}{a+b}\)

áp dụng BĐT Trê bư sép ta có:

\(\left(a^2+b^2+c^2\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\le3\left(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\right)=3VT\)

lại có a² + b² + c² \(\ge\) \(\frac{\left(a+b+c\right)^2}{3}\) nên:

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

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

sửa lại

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

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

áp dụng bđt cauchy ta có:

\(b^2+1\ge2b;c^2+1\ge2c;a^2+1\ge2a\)

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

\(=a+b+c-\frac{ab+bc+ca}{2}\)

áp dụng cauchy ta có:

\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)

\(\Rightarrow a+b+c-\frac{ab+bc+ca}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)

\(\Rightarrow\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\left(Q.E.D\right)\)

dấu bằng xảy ra khi a=b=c=1

đặt \(A=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}=a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\)

\(=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\right)\le3-\left(\frac{ab^2}{2b}+\frac{bc^2}{2c}+\frac{ca^2}{2a}\right)=3-\left(\frac{ab+bc+ca}{2}\right)\ge3-\frac{\left(a+b+c\right)^2}{6}=\frac{3}{2}\left(Q.E.D\right)\)

#)Giải :

Áp dụng BĐT Cauchy : \(\hept{\begin{cases}\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab}{2}\\\frac{b}{1+c^2}=b-\frac{bc^2}{1+c^2}\ge b-\frac{bc}{2}\\\frac{c}{1+a^2}=c-\frac{ca^2}{1+a^2}\ge c-\frac{ca}{2}\end{cases}}\)

\(\Rightarrow\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{1}{2}\left(ab+bc+ca\right)\ge3-\frac{1}{6}\left(a+b+c\right)^2=\frac{3}{2}\)

\(\Rightarrow\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\left(đpcm\right)\)

Theo BĐT AM-GM:

 \(\frac{a}{1+b^2}\)=a-\(\frac{ab^2}{1+b^2}\)\(\ge\)a-\(\frac{ab^2}{2b}\)=a-\(\frac{ab}{2}\)

Tương tự: \(\frac{b}{1+c^2}\)\(\ge\)b-\(\frac{bc}{2}\);\(\frac{c}{1+a^2}\)\(\ge\)c-\(\frac{ca}{2}\)

Suy ra \(\frac{a}{1+b^2}\)+\(\frac{b}{1+c^2}\)+\(\frac{c}{1+a^2}\)\(\ge\)a+b+c-\(\frac{1}{2}\)(ab+bc+ca)

Mặt khác thì theo BĐT AM-GM:9=a²+b²+c²+2(ab+bc+ca)

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

\(\Rightarrow\)\(\frac{1}{2}\)(ab+bc+ca)\(\le\)\(\frac{3}{2}\)

Cho nên  \(\frac{a}{1+b^2}\)+\(\frac{b}{1+c^2}\)+\(\frac{c}{1+a^2}\)\(\ge\)a+b+c-\(\frac{3}{2}\)=3-\(\frac{3}{2}\)=\(\frac{3}{2}\)

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

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

\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{9}{2}-3=\frac{3}{2}\)

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

\(=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1-3\)

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

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

\(=\frac{1}{2}[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)]\)\(\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

C/m BĐT phụ  \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\text{(*)   }\)  với x, y, z  dương

   Áp dụng BĐT Cô-si ta có:

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)

ÁP dụng  BĐT (*) ta có:

\(VT=\frac{1}{2}\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\)\(\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3\)

\(VT\ge\frac{1}{2}.9-3=\frac{3}{2}\left(đpcm\right)\)

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

>= \(\frac{\left(a+b+c\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}=\frac{a^2+b^2+c^2+2ab+2ac+2bc}{2ab+2ac+2bc}\)(bđt cauchy schawatz dạng engel)(1)

mà \(a^2+b^2+c^2>=ab+ac+bc\)

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

\(=\frac{3ab+3ac+3bc}{2ab+2ac+2bc}=\frac{3\left(ab+ac+bc\right)}{2\left(ab+ac+bc\right)}=\frac{3}{2}\)(2)

từ (1) và (2)\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>=\frac{3}{2}\)

dấu = xảy ra khi a=b=c

Mình dùng ''AM-GM ngược dấu'' như sau

Áp dụng bất đẳng thức AM-GM ta có \(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)

Tương tự với các phân thức khác rồi cộng vế theo vế ta được:

\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\left(\frac{ab}{2}+\frac{bc}{2}+\frac{ca}{2}\right)=3-\left(\frac{ab}{2}+\frac{bc}{2}+\frac{ca}{2}\right)\)

Mặt khác áp dụng bất đẳng thức AM-GM  \(9=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow ab+bc+ca\le\frac{3}{2}\)

Vậy \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge3-\frac{3}{2}=\frac{3}{2}\)

bạn ơi đoạn cuối áp dụng BĐT AM-GN  mk chưa hiểu lắm

À mình dùng như thế này nhá \(a^2+b^2+c^2\ge ab+bc+ca\left(1\right)\)

Bạn có thể chứng minh bằng tách đối xứng như sau

\(VT\left(1\right)=\left(\frac{a^2}{2}+\frac{b^2}{2}\right)+\left(\frac{b^2}{2}+\frac{c^2}{2}\right)+\left(\frac{c^2}{2}+\frac{a^2}{2}\right)\ge2\sqrt{\frac{a^2b^2}{4}}+2\sqrt{\frac{b^2c^2}{4}}+2\sqrt{\frac{c^2a^2}{4}}\)

\(=ab+bc+ca\)

Còn cách khác thì chứng minh tương đương

Bất đẳng thức(1) tương đương với \(\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)

Bất đẳng thức này hiển nhiên đúng nên suy ra (1) đúng

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

Tương tự ta có \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\); \(\frac{c}{1+a^2}\ge c-\frac{ac}{2}\)

\(\Rightarrow VT\ge a+b+c-\frac{1}{2}\left(ab+ac+bc\right)\ge3-\frac{1}{6}\left(a+b+c\right)^2=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(VT=\frac{4}{2.2\sqrt{a+b}}+\frac{4}{2.2\sqrt{b+c}}+\frac{4}{2.2\sqrt{c+a}}\)

\(VT\ge\frac{4}{a+b+4}+\frac{4}{b+c+4}+\frac{4}{c+a+4}\)

\(VT\ge\frac{36}{a+b+4+b+c+4+c+a+4}=\frac{36}{24}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=2\)