Cho a,b,c là các số thực dương thỏa mãn a+b+c=1.CMR:
$\frac{ab}{a^{2}+b^{2}}+\frac{bc}{b^{2}+c^{2}}+\frac{ca}{c^{2}+a^{2}}$ +$\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$\geq \frac{15}{4}$
đề này hình như thấy ở đâu rồi –  thuhuong1607hhpt 29-08-15 10:35 PM
http://diendantoanhoc.net/forum/topic/145615-fracaba2b2-geq-frac154/

Làm 1 lần rồi không muốn tìm lại.
Ta có:
     $\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{1}{4}(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
$=\frac{1}{4}[3+\sum(\frac{a}{b}+\frac{b}{a})]=\frac{1}{4}[3+\sum(\frac{a^2+b^2}{ab})]$
Theo bất đẳng thức AM - GM, ta có:
     $\sum (\frac{ab}{a^2+b^2})+\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\sum (\frac{ab}{a^2+b^2})+\frac{1}{4}[3+\sum(\frac{a^2+b^2}{ab})]$
 $= \sum(\frac{ab}{a^2+b^2}+\frac{1}{4}.\frac{a^2+b^2}{ab})+\frac{3}{4}$
 $\geq 1+1+1+\frac{3}{4}=\frac{15}{4}$
Đẳng thức xảy ra tại $a=b=c=\frac{1}{3}.$
có $\frac{1}{a}=\frac{a+b+c}{a}=1+\frac{b}{a}+\frac{c}{a}$
tương tự có $\frac{1}{b}=1+\frac{a}{b}+\frac{c}{b}$
                   $\frac{1}{c}=1+\frac{a}{c}+\frac{b}{c}$
$\frac{ab}{a^{2}+b^{2}}+\frac{bc}{b^{2}+c^{2}}+\frac{ca}{c^{2}+a^{2}}=\frac{1}{\frac{a}{b}+\frac{b}{a}}+\frac{1}{\frac{b}{c}+\frac{c}{b}}+\frac{1}{\frac{c}{a}+\frac{a}{c}}\geq\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3}$
ta cần chứng minh $\frac{9}{t-3}+\frac{t}{4}\geq\frac{15}{4}$  (*) với $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
thật vậy (*) $\Leftrightarrow(t-9)^{2}\geq0$ 
"=" xảy ra khi $a=b=c=\frac{1}{3}$

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