Cho a,b,c >0. CMR:
$\frac{ab}{a+2a^2+a^3+2b^4+2c^8+10}+\frac{bc}{b+2b^2+b^3+2c^4+2a^8+10}+\frac{ca}{c+2c^2+c^3+2a^4+2b^8+10}< \frac{1}{4}$
\[\sum {\frac{{ab}}{{a + 2{a^2} + {a^3} + 2{b^4} + 2{c^8} + 10}} < \frac{1}{4}} \]
Started By mango, 23-05-2012 - 13:32
#1
Posted 23-05-2012 - 13:32
- cool hunter and Dung Dang Do like this
#2
Posted 29-05-2012 - 19:42
Giải như sau:Cho a,b,c >0. CMR:
$\frac{ab}{a+2a^2+a^3+2b^4+2c^8+10}+\frac{bc}{b+2b^2+b^3+2c^4+2a^8+10}+\frac{ca}{c+2c^2+c^3+2a^4+2b^8+10}< \frac{1}{4}$
$$\sum{\frac{ab}{a+2a^2+a^3+2b^4+2c^8+10}}=\sum{\frac{ab}{(a+a^3+a^2+a^2)+(2b^4+2)+(c^8+1+1+1)+(c^8+5)}}$$
$$<\sum{\frac{ab}{4\sqrt[4]{a.a^3.a^2.a^2}+2\sqrt{2b^4.2}+4\sqrt[4]{c^8}}}=\sum{\frac{ab}{4a^2+4b^2+4c^2}}\le \sum{\frac{ab}{4(ab+bc+ca)}}=\frac{1}{4}$$
Do đó có $đpcm$
Edited by nguyenta98, 29-05-2012 - 19:43.
- cool hunter, mango and 013 like this
#3
Posted 16-12-2016 - 20:13
∑aba+2a2+a3+2b4+2c8+10=∑ab(a+a3+a2+a2)+(2b4+2)+(c8+1+1+1)+(c8+5)∑aba+2a2+a3+2b4+2c8+10=∑ab(a+a3+a2+a2)+(2b4+2)+(c8+1+1+1)+(c8+5)
<∑ab44√a.a3.a2.a2+2√2b4.2+44√c8=∑ab4a2+4b2+4c2≤∑ab4(ab+bc+ca)=14<∑ab4a.a3.a2.a24+22b4.2+4c84=∑ab4a2+4b2+4c2≤∑ab4(ab+bc+ca)=14
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