ad
\frac{1}{x} +\frac{1}{y} \geq \frac{4}{x+y} ta có
\frac{1}{4a+ b+c} \leq \frac{1}{4} (\frac{1}{2a +b} +\frac{1}{2a +c})
có \frac{1}{2a +b} \leq \frac{1}{9}(\frac{1}{a} +\frac{1}{a} +\frac{1}{b})
TT \Rightarrow \frac{1}{4a +b +c} \leq \frac{1}{36} (\frac{4}{a} +\frac{1}{b} +\frac{1}{c})
Từ đó \Rightarrow VT \leq \frac{1}{6} (\frac{1}{a} + \frac{1}{b} +\frac{1}{c})
ta cần cm \frac{1}{a}+ \frac{1}{b} +\frac{1}{c} \leq 1
ta thấy 3(\frac{1}{a^{2}} + \frac{1}{b^{2}}+ \frac{1}{c^{2}}) \geq (\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c})^{2}
\Leftrightarrow 4( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c})^{2} \leq \frac{1}{a} +\frac{1}{b} +\frac{1}{c} +3 (do gt)
\Leftrightarrow \frac{1}{a} +\frac{1}{b} +\frac{1}{c} \leq 1
\Rightarrow đpcm
dấu "=" \Leftrightarrow a=b=c=3