Ô, giờ mới 10 giờ kém

Bóng bầu dục

http://diendantoanhoc.net/topic/104095-cmr-a2b2c2abc-geq-4/

vô đó,,ta sẽ có 3+abc>=4   => abc>=1          =>-abc<1             => dùng vs cái trên => dpcm

Trên tròi

áp dụng bđt phụ

a²+b²+c²>= 1/3(a+b+c)²  

=> a+b+c <=3  

chịu thôi

cai cuoi la 1,8

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}=\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{1}{1}-\dfrac{1}{6}=\dfrac{6}{6}-\dfrac{1}{6}=\dfrac{5}{6}\)