cho a>0,b>0,c>0
cmr \(\frac{2ab}{\left(c+a\right)\left(c+b\right)}+\frac{2bc}{\left(a+b\right)\left(a+c\right)}+\frac{2ca}{\left(b+a\right)\left(b+c\right)}\ge\frac{5}{3}\)
dấu "=" xảy ra khi nào
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3 tháng 8 2017
ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3
AT
23 tháng 1 2017
Ta có: \(a+b+c=\frac{1}{2}\) \(\Rightarrow\left\{\begin{matrix}a=\frac{1}{2}-b-c\\b=\frac{1}{2}-a-c\\c=\frac{1}{2}-a-b\end{matrix}\right.\) hay \(\left\{\begin{matrix}a+b=\frac{1}{2}-c\\b+c=\frac{1}{2}\\a+c=\frac{1}{2}-b\end{matrix}\right.\)
\(P=\frac{\left(2ab+c\right)\left(2bc+a\right)\left(2ac+b\right)}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}\)
\(=\frac{\left(2ab+\frac{1}{2}-a-b\right)\left(2bc+\frac{1}{2}-b-c\right)\left(2ca+\frac{1}{2}-a-c\right)}{\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)}\)
\(=\frac{2\left(ab+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}b\right).2\left(bc+\frac{1}{4}-\frac{1}{2}b-\frac{1}{2}c\right).2\left(ca+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}c\right)}{\left(ac+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}c\right)\left(bc+\frac{1}{4}-\frac{1}{2}b-\frac{1}{2}c\right)\left(ab+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}b\right)}\)
\(=2.2.2=8\)
Vậy với \(a+b+c=\frac{1}{2}\) và \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ne0\) thì \(P=8\)
\(\frac{2ab}{\left(c+a\right)\left(c+b\right)}+\frac{2bc}{\left(a+b\right)\left(a+c\right)}+\frac{2ca}{\left(b+a\right)\left(b+c\right)}\ge\frac{3}{2}\) thì phải