cho a, b, c>0
CM: $\sum \frac{2a}{a^6+b^4}\leq \sum \frac{1}{a^4}$
$\sum \frac{2a}{a^{6}+b^{4}}\leq \sum \frac{2a}{2a^{3}b^{^{2}}}=\sum \frac{1}{a^{2}b^{2}}\leq \sum \frac{1}{a^{4}}$
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$$\sum \frac{2a}{a^6+b^4}\leq \sum \frac{1}{a^4}$$
$$\frac{1}{a^{4}}+ \frac{1}{b^{4}}= \frac{a^{2}}{a^{6}}+ \frac{1}{b^{4}}\geq \frac{\left ( a+ 1 \right )^{2}}{a^{6}+ b^{4}}\geq \frac{4a}{a^{6}+ b^{4}}$$
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