B3
Đặt S=$\sum \dfrac{a}{\sqrt[3]{a+2b}}$
P=$\sum a(a+2b)$
Áp dung BDT holder cho 4cặp số $\Rightarrow S^{3}P\geq (a+b+c+d)^{4}\Rightarrow S^{3}\geq \dfrac{(a+b+c+d)^{4}}{\sum a(a+2b)}= (a+b+c+d)^{2}= 1\Rightarrow S\geq 1$ (dpcm)
Cách khác:
$\sum \dfrac{a}{\sqrt[3]{a+2b}}\geq \dfrac{3a}{a+2b+2}\geq \dfrac{3(a+b+c)^2}{a^2+b^2+c^2+2(ab+ac+bc)+2(a+b+c)}=1$
Cách khác:
Ta có $\sqrt[3]{1.1.(a+2b)}\leq \dfrac{a+2b+2}{3}$
$VT=\sum \dfrac{a^{2}}{\sqrt[3]{a+2b}}\geq \sum \dfrac{3a^{2}}{a+2b+2}=3(\sum \dfrac{a^{2}}{a^{2}+2ab+2a})\geq 3\dfrac{(a+b+c)^{2}}{(a+b+c)^{2}+2(a+b+c)}=3.\dfrac{1}{3}=1.$
Dấu $"="$ xảy ra $\Leftrightarrow a=b=c=\dfrac{1}{3}$