Cho \(a\ge1,b\ge1\)
Cm: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)
cm với \(a\ge b\ge1:\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)
Ta có: \(a\ge b\Rightarrow1+b^2\le1+a^2\)
\(\Rightarrow\frac{1}{1+b^2}\ge\frac{1}{1+a^2}\Rightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{1}{1+a^2}+\frac{1}{1+a^2}\)
\(\Leftrightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+a^2}\)
Cho hai số a,b thỏa mãn: \(a\ge1,b\ge1\). CMR: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)
\(\Leftrightarrow\left(2+a^2+b^2\right)\left(1+ab\right)\ge2\left(1+a^2\right)\left(1+b^2\right)\)
\(\Leftrightarrow2+2ab+a^2+b^2+ab\left(a^2+b^2\right)\ge2+2a^2+2b^2+2a^2b^2\)
\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)
\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\) (luôn đúng với mọi \(a\ge1;b\ge1\))
Cách khác:
\(\Leftrightarrow\left(\frac{1}{1+a^2}-\frac{1}{1+ab}\right)+\left(\frac{1}{1+b^2}-\frac{1}{1+ab}\right)\ge0\)
\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)\left[b\left(1+a^2\right)-a\left(1+b^2\right)\right]}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\) (luôn đúng).
Cho \(a\ge1,b\ge1,c\ge1\) CMR
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge\frac{3}{abc+1}\)
khó quá bạn ơi mình cần thêm thời gian để làm
cho a,b>0 cm\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\) nếu \(ab\ge1\)
b) cho a,b,c\(\ge\)1. CMR \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)
\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)
\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)
\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)
b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)
\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)
Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)
Cộng vế với vế ta có đpcm
giúp cái CM BĐT; \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{ab+1}\)với \(ab\ge1\)
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{ab+1}\)
\(\Leftrightarrow\left(\frac{1}{1+a^2}-\frac{1}{ab+1}\right)+\left(\frac{1}{1+b^2}-\frac{1}{1+ab}\right)\ge0\)
\(\Leftrightarrow\frac{ab-a^2}{\left(1+a^2\right)\left(ab+1\right)}+\frac{ab-b^2}{\left(1+b^2\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(ab+1\right)}+\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)}{ab+1}\left(\frac{b}{1+b^2}-\frac{a}{1+a^2}\right)\ge0\)
\(\Leftrightarrow\frac{a-b}{ab+1}.\frac{b+ba^2-a-ab^2}{\left(1+a^2\right)\left(1+b^2\right)}\ge0\)
\(\Leftrightarrow\frac{a-b}{ab+1}.\frac{ab\left(a-b\right)-\left(a-b\right)}{\left(1+a^2\right)\left(1+b^2\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(ab+1\right)\left(1+a^2\right)\left(1+b^2\right)}\ge0\)
Vì \(ab\ge1\) nên BĐT trên luôn đúng.
Vậy bđt ban đầu dc chứng minh .
Cho a,b,c \(\ge\)0 TM \(a^2+b^2+c^2=1\) CM
\(\frac{c}{1+ab}+\frac{b}{1+ac}+\frac{a}{1+bc}\ge1\)
Cho \(ab\ge1\). Chứng minh rằng: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)
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
a) Cho a+b+c=0 và abc khác 0, Tính
P=\(\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+b^2-c^2}+\frac{1}{a^2+b^2-c^2}\)
b) Cho 2 số a và b thỏa mãn \(a\ge1;b\ge1\). Chứng minh \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)
Cứu vs !!
\(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=b^2+2bc+c^2\\b^2=a^2+2ac+c^2\\c^2=a^2+2ab+b^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b^2+c^2-a^2=-2bc\\a^2+c^2-b^2=-2ac\\a^2+b^2-c^2=-2ab\end{matrix}\right.\Rightarrow P=\frac{1}{-2bc}+\frac{1}{-2ac}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=0\)
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
\(\Rightarrow P=\frac{1}{a^2-2bc-a^2}+\frac{1}{c^2-2ab-c^2}+\frac{1}{b^2-2ac-b^2}\)
\(P=\frac{1}{-2bc}+\frac{1}{-2ab}+\frac{1}{-2ac}\)
\(P=\frac{a+b+c}{-2abc}=\frac{0}{-2abc}=0\)
\(xét:\frac{1}{a^2+1}+\frac{1}{b^2+1}-\frac{2}{1+ab}=\left(\frac{1}{a^2+1}-\frac{1}{1+ab}\right)+\left(\frac{1}{b^2+1}-\frac{1}{1+ab}\right)=\frac{1+ab-a^2-1}{\left(a^2+1\right)\left(1+ab\right)}+\frac{1+ab-1-b^2}{\left(b^2+1\right)\left(1+ab\right)}=\frac{a\left(b-a\right)}{\left(a^2+1\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(b^2+1\right)\left(1+ab\right)}=\left(a-b\right)\left(\frac{b}{\left(b^2+1\right)\left(1+ab\right)}-\frac{a}{\left(a^2+1\right)\left(1+ab\right)}\right)=\left(a-b\right)\left(\frac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(ab+1\right)\left(b^2+1\right)}\right)=\left(a-b\right)\left(\frac{\left(ab-1\right)\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\) \(\left(a-b\right)^2\frac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\left(do:a\ge1;b\ge1\right)\Rightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\left(a\ge1;b\ge1\right)\)
Chứng minh: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\) với a,b\(\ge1\)
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 \(a=b\)