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
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