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C

chaizo1234567

Bài làm

Ta có:
A=c2a2+b2+4c22ab+c2ab+ab+c2c2A=\frac{c^2}{a^2+b^2}+\frac{4c^2}{2ab}+\frac{c^2}{ab}+\frac{ab+c^2}{c^2}
Do a+b\leqc
\Rightarrowc2c^2\geqa2+b2+2aba^2+b^2+2ab
\RightarrowA\geq(c+2c+c)2a2+b2+3ab\frac{(c+2c+c)^2}{a^2+b^2+3ab}+a2+b2+3abc2\frac{a^2+b^2+3ab}{c^2}
\RightarrowA\geq8.................
 
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