Question

Answer 1

Bài 12:
Áp dụng BĐT Cauchy-Schwarz:

$\frac{1}{a}+\frac{1}{4b}=\frac{2^2}{4a}+\frac{1^2}{4b}\geq \frac{(2+1)^2}{4a+4b}=\frac{9}{4(a+b)}=\frac{9}{4}$

Vậy ta có đpcm.
Dấu "=" xảy ra khi $\frac{1}{a}=\frac{1}{2b}\Rightarrow a=\frac{2}{3}; b=\frac{1}{3}$

 

Answer 2

Bài 13:
Áp dụng BĐT Cauchy-Schwarz:

$\frac{4}{2a+b+c}\leq \frac{4}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

$\frac{4}{a+2b+c}\leq \frac{4}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)$

$\frac{4}{a+b+2c}\leq \frac{4}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)$

Cộng các BĐT trên thì:

$\sum \frac{4}{2a+b+c}\leq \sum \frac{1}{a}$
Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$