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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)
xét hiệu \(\frac{1}{1+a^2}-\frac{1}{1+ab}+\frac{1}{1+b^2}-\frac{1}{1+ab}\)
quy đồng làm nốt nha
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 \(a=b\)
\(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=b^2+2bc+c^2\\b^2=a^2+2ac+c^2\\c^2=a^2+2ab+b^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b^2+c^2-a^2=-2bc\\a^2+c^2-b^2=-2ac\\a^2+b^2-c^2=-2ab\end{matrix}\right.\Rightarrow P=\frac{1}{-2bc}+\frac{1}{-2ac}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=0\)
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
\(\Rightarrow P=\frac{1}{a^2-2bc-a^2}+\frac{1}{c^2-2ab-c^2}+\frac{1}{b^2-2ac-b^2}\)
\(P=\frac{1}{-2bc}+\frac{1}{-2ab}+\frac{1}{-2ac}\)
\(P=\frac{a+b+c}{-2abc}=\frac{0}{-2abc}=0\)
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 !
https://hoc247.net/hoi-dap/toan-10/tim-gtnn-cua-b-1-a-2-b-2-1-ab-4ab-faq164975.html
Chứng minh \(\frac{2}{1+bc}-\frac{1}{1+b^2}-\frac{1}{1+c^2}\le0\)
Cảm ơn bạn nhé, mk vt thiếu đề
Dùng phương pháp biến đổi tương đương nhé!!!
Ta có : \(\dfrac{1}{1+a^2}\) + \(\dfrac{1}{1+b^2}\) \(\ge\) \(\dfrac{2}{1+ab}\)
<=>( \(\dfrac{1}{1+a^2}\) - \(\dfrac{1}{1+ab}\) ) + ( \(\dfrac{1}{1+b^2}\) - \(\dfrac{1}{1+ab}\) ) \(\ge\) 0
<=> \(\dfrac{1+ab-1-a^2}{\left(1+a^2\right)\left(1+ab\right)}\) + \(\dfrac{1+ab-1-b^2}{\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0
<=> \(\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}\) + \(\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0
<=> \(\dfrac{a\left(b-a\right)\left(1+b^2\right)+b\left(a-b\right)\left(1+a^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\) \(\ge\) 0
<=> \(a\left(b-a\right)\left(1+b^2\right)-b\left(b-a\right)\left(1+a^2\right)\) \(\ge\) 0
<=> \(\left(b-a\right)\left(a+ab^2-b-a^2b\right)\) \(\ge\) 0
<=> \(\left(b-a\right)\left[ab\left(b-a\right)-\left(b-a\right)\right]\) \(\ge\) 0
<=> \(\left(b-a\right)\left(b-a\right)\left(ab-1\right)\) \(\ge\) 0
<=> \(\left(b-a\right)^2\left(ab-1\right)\) \(\ge\) 0 (1)
Mà \(\left\{{}\begin{matrix}\left(b-a\right)^2\ge0\\ab-1\ge0\end{matrix}\right.\) ( vì ab \(\ge\)1)
=> \(\left(b-a\right)^2\left(ab-1\right)\) \(\ge\) 0
=> (1) luôn đúng
Vậy đpcm ....
Ta có: \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)
\(\Leftrightarrow\left(\dfrac{1}{1+a^2}-\dfrac{1}{1+b^2}\right)+\left(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\right)\ge0\)
\(\Leftrightarrow\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)
BĐT cuối cùng đúng vì \(a.b\ge1\Rightarrowđpcm\)