Ta có : \(\left(a+b\right)^2\ge4ab\Leftrightarrow\frac{ab}{a+b}\le\frac{a+b}{4}\)
Tương tự : \(\frac{bc}{b+c}\le\frac{b+c}{4}\) ; \(\frac{ac}{a+c}\le\frac{a+c}{4}\)
Cộng các bđt trên theo vế được :
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{c+a}\le\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}\)
Cho \(a,b,c>0\) . CMR :
\(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
cho 3 số a,b,c thỏa mãn điều kiện \(\frac{1}{bc-a^2}+\frac{1}{ca-b^2}+\frac{1}{ab-c^2}=0\)
CMR: \(\frac{a}{\left(bc-a^2\right)^2}+\frac{b}{\left(ca-b^2\right)^2}+\frac{c}{\left(ab-c^2\right)^2}=0\)
cho a, b, c >0 thỏa mãn \(a^2+b^2+c^2=1\)., Chứng minh rằng \(\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ac}\le\frac{9}{2}\)
CMR nếu
\(c^2+2\left(ab-ac-bc\right)=0,b\ne c,a+b\ne c\) thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
Cho a,b,c > 0. Chứng minh: a) \(\frac{ab}{c}\) +\(\frac{bc}{a}\) \(\ge2b\)
b) \(\frac{ab}{c}\) + \(\frac{bc}{a}\) + \(\frac{ac}{b}\) \(\ge\) \(a+b+c\)
Thực hiện phép tính :
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
cho a+b+a=10 chứng minh \(\frac{a^2}{bc}+\frac{b}{ac}^2+\frac{c^2}{ab}\)
CMR với a,b,c > 0 thì \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
Cho \(\frac{a}{b+c}\)+\(\frac{b}{c+a}\)+\(\frac{c}{a+b}\)=1.CMR: \(\frac{a^2}{b+c}\)+\(\frac{b^2}{c+a}\)+\(\frac{c^2}{a+b}\)=0
