Ta bien doi BDT can chung minh
\(a+b\ge\frac{4ab}{1+ab}\)
\(\Leftrightarrow a+a^2b+b+ab^2\ge4ab\)
\(\Leftrightarrow a+\frac{1}{a}+b+\frac{1}{b}\ge4\)
Ta co:
\(a+\frac{1}{a}\ge2\)
\(b+\frac{1}{b}\ge2\)
\(\Rightarrow a+\frac{1}{a}+b+\frac{1}{b}\ge4\)
Dau '=' xay ra khi \(a=b=1\)
Cho a,b>0 tm a+b=4ab Cm \(\frac{a}{4b^2+1}+\frac{b}{4a^2+1}\ge\frac{1}{2}\)
Cho a,b>0.Chứng minh
\(a+b\ge\frac{4ab}{1+ab}\)
a>0, b>0. \(CM:a+b\ge\frac{4ab}{1+ab}\)
cho a,b>0 thỏa mãn a+b=4ab. CMR
\(\frac{a}{4b^2+1}+\frac{b}{4a^2+1}\ge\frac{1}{2}\)
Cho a;b >0 thỏa mãn : \(12\ge\left(a+b\right)^3+4ab\). CMR:
\(\frac{1}{a+1}+\frac{1}{b+1}+2017ab\le2018\)
cho \(0< a\le\frac{1}{2},0< b\le\frac{1}{2}.CM:\left(\frac{a+b}{2-a-b}\right)^2\ge\frac{ab}{\left(1-a\right)\left(1-b\right)}\)
Cho a,b >0 và a+b= 1. Chứng minh : S = \(\frac{1}{a^2+b^2}+\frac{1}{ab}+4ab\ge7\)
Cho a,b,c>0 CM: \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)
C/m: a,b lớn hơn 0 thì (a+b)(ab+1) \(\ge\) 4ab
