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

cosi la ra

1. Áp dụng BĐT Cauchy dạng Engle, ta có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

\(\frac{1}{3}\left(a^3+b^3+a+b\right)+ab\le a^2+b^2+1\)

\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2+1-ab\right)+ab\le a^2+b^2+1\)

\(\Leftrightarrow\left(a^2+b^2+1\right)\left(\frac{a+b}{3}-1\right)-ab\left(\frac{a+b}{3}-1\right)\le0\)

\(\Leftrightarrow\left(a^2+b^2+1-ab\right)\left(\frac{a+b}{3}-1\right)\le0\)

Vì a, b dương \(\Rightarrow a^2+b^2+1-ab>0\Rightarrow\left(\frac{a+b}{3}-1\right)\le0\Leftrightarrow a+b\le3\)

\(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=2a+2b+\frac{8}{a}+\frac{2}{b}-\left(a+b\right)\ge8+4-3=9\)

Áp dụng BĐT Cauchy cho a ; b dương

Dấu "=" xảy ra \(\Leftrightarrow a=2;b=1\)

Áp dụng BĐT AM-GM ta có:

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

\(=\left(a+1\right)-\frac{ab+b}{2}\). Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT\ge3+\left(a+b+c\right)-\frac{ab+bc+ca+a+b+c}{2}\)

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

Dấu "=" <=> \(a=b=c=1\)

\(Áp dụng BĐT AM-GM ta có: \(\frac{a+1}{b^2+1}=\left(a+1\right)-\frac{b^2\left(a+1\right)}{b^2+1}\ge\left(a+1\right)-\frac{b^2\left(a+1\right)}{2b}\) \(=\left(a+1\right)-\frac{ab+b}{2}\). Tương tự cho 2 BĐT còn lại rồi cộng theo vế: \(VT\ge3+\left(a+b+c\right)-\frac{ab+bc+ca+a+b+c}{2}\) \(\ge3+\left(a+b+c\right)-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\) Dấu "=" <=> \(a=b=c=1\)\)

\(\frac{1}{a^2}=\frac{1}{\left(bc\right)^2}\)

\(\Rightarrow\frac{1}{a^2}+1=\frac{1}{\left(bc\right)^2}+1\ge2\frac{1}{bc}=2a\)

Bạn Hoàng sai rồi nhé: 

cho \(a=\frac{3}{2};b=2;c=\frac{1}{3}\) (t/m đk abc=1)

Suy ra \(a+b+c=\frac{3}{2}+2+\frac{1}{3}=3,8\left(3\right)>3\) nhé

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

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

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

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

\(=4+2\frac{a}{b}+2\frac{b}{a}+1\)

\(=5+2\left(\frac{a}{b}+\frac{b}{a}\right)\)\(\ge5+2.2=9\)

c/m:  \(\frac{a}{b}+\frac{b}{a}\ge2\) với a,b dương

Dấu "=" xảy ra  \(\Leftrightarrow\)\(a=b\)

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

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

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

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

\(=4+2\frac{b}{a}+2\frac{a}{b}+1\)

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

\(=5+2\left(\frac{a}{b}+\frac{b}{a}\right)\)(1)

.Ta cần chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\)

Không mất tính tổng quát, giả sử a, b \(>0\)va \(a\ge b\). \(m\ge0\). Có thể viết \(a=b+m\)

Vậy \(\frac{a}{b}+\frac{b}{a}=\frac{b+m}{b}+\frac{b}{b+m}=1+\frac{m}{b}+\frac{b}{b+m}\ge1+\frac{m}{b+m}+\frac{b}{b+m}\)

\(=1+\frac{m+b}{b+m}=1+1=2\)

Từ đây, ta suy ra được \(\left(1\right)\ge5+2.2=9^{\left(đpcm\right)}\)

Dấu = xảy ra khi a = b (m = 0)