Chứng minh BĐT

K

kakashi_hatake

$(1+\dfrac{1}{a})(1+\dfrac{1}{b})(1+\dfrac{1}{c})=\dfrac{(2a+b+c)(a+2b+c)(a+b+2c)}{abc} \\ a+b+2c=(a+c)+(b+c) \ge 2.\sqrt{ac}+2\sqrt{bc} \ge 4.\sqrt[4]{abc^2} \\ \rightarrow (2a+b+c)(a+2b+c)(a+b+2c) \ge 64.\sqrt[4]{a^4b^4c^4}=64abc$

Thay vào đc đpcm
 
C

congchuaanhsang

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