toán!!!!pro!!!!!!!!!BĐT

B

bigbang195

Bai nay quen thuoc ma` :
[TEX]\sum a \ge \sum a^2b^2 \Leftrightarrow 2\sum a \ge 2\sum a^2b^2[/TEX]
[TEX]\Leftrightarrow 2\sum a+\sum a^4 \ge 2\sum a^2b^2 +\sum a^4 =(\sum a^2)^2=9[/TEX]
ma` [TEX]2\sum a+\sum a^4 = \sum a+\sum a+\sum a^4 \geq^{AM-GM} 3\sum \sqrt[3]{a^6} =3\sum a^2 =3.3 =9[/TEX]
 
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