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