\(\Rightarrow\frac{a+b+c}{abc}=\frac{1}{9}\Leftrightarrow\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=\frac{2}{9}\)
Lại có \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=1\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=1\)
Vậy 1/a^2+1/b^2+1/c^2=1-2/9=7/9 ( Sê đài )
cho a,b,c>0.CMR:\(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
cho a,b,c>0 . Cmr: \(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)
(sử dụng AM-GM)
Cho a,b,c >0, a+b+c=3. CMR: \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\)
Cho a, b, c > 0 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
CMR : \(\frac{a^2}{a+bc}+\frac{b^2}{b+ca}+\frac{c^2}{c+ab}\) ≥ \(\frac{a+b+c}{4}\)
Cho a,b,c>0, abc\(=\)1
CMR: \(\frac{1}{a^2-ab+b^2}+\frac{1}{b^2-bc+c^2}+\frac{1}{c^2-ca+a^2}\le a+b+c\)
Cho a,b,c > 0 thỏa mãn \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{1}{2}\) . CMR:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Cho \(\left(a+b+c\right)^2=a^2+b^2+c^2\) và a,b,c là 3 số khác 0 .CMR : \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
1) Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) . Tính giá trị biểu thức: D= \(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}\)
2) Cho a+b+c=\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\); abc khác 0. Ch/m \(a^6+b^6+c^6=3a^2b^2c^2\)
Cho a,b,c là 3 số nguyên khác 0 thỏa mãn:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\).CMR:\(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\)là số chính phương
