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

\(\Leftrightarrow\frac{1}{\left(1+a^2\right)}+\frac{1}{\left(1+b^2\right)}-\frac{2}{\left(1+ab\right)}>=0\)

\(\Leftrightarrow\left[\frac{1}{\left(1+a^2\right)}-\frac{1}{\left(1+ab\right)}\right]+\left[\frac{1}{\left(1+b^2\right)}-\frac{1}{\left(1+ab\right)}\right]>=0\)

\(\Leftrightarrow\left[\frac{a\left(b-c\right)}{\left(1+a^2\right)\left(1+ab\right)}\right]+\left[\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\right]>=0\)

\(\frac{\left[a\left(b-a\right)\left(1+b^2\right)-b\left(b-a\right)\left(1+a^2\right)\right]}{\left[\left(1+a^2\right)\left(1+b^2\right)\left(1+b^2\right)\left(1+ab\right)^2\right]}>=0\)

\(\frac{\left[\left(b-a\right)\left(a+ab^2-b+ba^2\right)\right]}{\left[\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)^2\right]}>=0\)

\(\frac{\left[\left(b-a\right)\left[\left(a-b\right)+ab\left(b-a\right)\right]\right]}{\left[\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)^2\right]}>=0\)

\(\frac{\left[\left(b-a\right)^2\left(ab-1\right)\right]}{\left[\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)^2\right]}>=0\left(1\right)\)

Mẫu số luôn lớn hơn 1 

\(\left(b-a\right)^2>=0\)  voi moi a,b

Vì a,b >=1 nên ( ab-1) > = 0

​Nên (1)  dụng

Tu "dung"doi thanh dung

Ta có: \(\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}=\left(1+ab\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\)

mà \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+a^2+b^2}\)( Áp dụng BĐT phụ \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\))

Mặt khác: \(a^2+b^2\ge2ab\)

=> \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+2ab}=\frac{2}{1+ab}\)

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

Bạn cần biết  \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)  (nếu bạn chưa biết thì xét hiệu) 

Ta có: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\)

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

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

\(\ge\frac{4}{2ab+2}=\frac{2}{ab+1}\)

Dấu "=" xảy ra khi \(a=b\)

Nếu không áp dụng BĐT thì chuyển vế cũng được nhưng hơi dài :

Mình thử làm thôi nhé :

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

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

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

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

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

....

Giải bất mà không được dùng bất ? Vô lý thế ??

Bài Đạt chưa làm hết,mình làm nốt nha !

1,

\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)

\(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=b^2+2bc+c^2\\b^2=a^2+2ac+c^2\\c^2=a^2+2ab+b^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b^2+c^2-a^2=-2bc\\a^2+c^2-b^2=-2ac\\a^2+b^2-c^2=-2ab\end{matrix}\right.\Rightarrow P=\frac{1}{-2bc}+\frac{1}{-2ac}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=0\)

a) \(P=\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+b^2-c^2}+\frac{1}{a^2+c^2-b^2}\) ( Sửa đề )

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

Vì a + b + c = 0

Nên a + b = -c

=> ( a + b )² = (-c)² = c²

Tương tự: ( b + c )² = a² và ( a + c )² = b²

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

\(P=\frac{1}{-2bc}+\frac{1}{-2ab}+\frac{1}{-2ac}\)

\(P=\frac{a+b+c}{-2abc}=\frac{0}{-2abc}=0\)

\(xét:\frac{1}{a^2+1}+\frac{1}{b^2+1}-\frac{2}{1+ab}=\left(\frac{1}{a^2+1}-\frac{1}{1+ab}\right)+\left(\frac{1}{b^2+1}-\frac{1}{1+ab}\right)=\frac{1+ab-a^2-1}{\left(a^2+1\right)\left(1+ab\right)}+\frac{1+ab-1-b^2}{\left(b^2+1\right)\left(1+ab\right)}=\frac{a\left(b-a\right)}{\left(a^2+1\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(b^2+1\right)\left(1+ab\right)}=\left(a-b\right)\left(\frac{b}{\left(b^2+1\right)\left(1+ab\right)}-\frac{a}{\left(a^2+1\right)\left(1+ab\right)}\right)=\left(a-b\right)\left(\frac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(ab+1\right)\left(b^2+1\right)}\right)=\left(a-b\right)\left(\frac{\left(ab-1\right)\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\) \(\left(a-b\right)^2\frac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\left(do:a\ge1;b\ge1\right)\Rightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\left(a\ge1;b\ge1\right)\)