3 replies to this topic

[leduylinh1998]

Cho a,b,c>0. CMR:

$\sum \frac{2a}{a^{6}+b^{4}}\leq \sum \frac{1}{a^{4}}$

[kfcchicken98]

có $\frac{1}{a^{4}}= \frac{a^{2}}{a^{6}}$
$\frac{a^{2}}{a^{6}}+ \frac{1}{b^{4}} \geq \frac{(a+1)^{2}}{a^{6}+b^{4}}\geq \frac{4a}{a^{6}+b^{4}}$
tương tự, suy ra $2\sum \frac{1}{a^{4}} \geq 4\sum \frac{a}{a^{6}+b^{4}}$ (dpcm)

[hoctrocuanewton]

áp dụng bđt AM-GM 

ta có

$\sum \frac{2a}{a^{6}+b^{4}}\leq \sum \frac{2a}{2a^{3}b^{2}}= \sum \frac{1}{a^{2}b^{2}}$

mà $\sum \frac{1}{a^{2}b^{2}}\leq \sum \frac{1}{a^{4}}$

vậy được đpcm

[leduylinh1998]

Ta có:$\sum \frac{2a}{a^{6}+b^{4}}\leq\sum \frac{2a}{2a^{3}b^{2}}=\sum \frac{1}{a^{2}b^{2}}\leq \sum \frac{1}{a^{4}}$