TA CÓ :$\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ ($B-C-S$)Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$
TA CÓ : BĐT $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ (BĐT $B-C-S$)Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$
TA CÓ :$\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
$\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$
($B-C-S$)Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$