[不等式] $\sum2a^2/(c^2+3ac+2a^2)\ge1$

Let $a,b,c\in R^+$,prove that
\[\sum\frac{2a^2}{c^2+3ac+2a^2}\ge1\]

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proof:
\[ \frac{2a^{2}}{2a^{2}+3ac+c^{2}}=\frac{2}{2+3\frac{c}{a}+\frac{c^{2}}{a^{2}}} \]
if we let $ u=\frac{c}{a},v=\frac{a}{b},w=\frac{b}{c} $,we get the result: $uvw=1$
and the original inequality become:
\[ \sum{\frac{2}{2+3u+u^{2}}}\geq 1 \]
Then,note that: $ u=\frac{yz}{x^{2}},v=\frac{xz}{y^{2}},w=\frac{xy}{z^{2}} $
Our inequality become:
\[ \sum{\frac{2x^{4}}{2x^{4}+3x^{2}yz+y^{2}z^{2}}}\geq 1 \]
Then,by Cauchy-Schwarz:
\[ [ \sum{\frac{2x^{4}}{2x^{4}+3x^{2}yz+y^{2}z^{2}}}\geq \frac{2(x^{2}+y^{2}+z^{2})^{2}}{2\sum{x^{4}}+3\sum{x^{2}yz}+\sum{y^{2}z^{2}}} \]
Thus.it suffice to prove that:
\[  \frac{2(x^{2}+y^{2}+z^{2})^{2}}{2\sum{x^{4}}+3\sum{x^{2}yz}+\sum{y^{2}z^{2}}}\geq 1 \]
Or
\[ \sum{x^{2}y^{2}}\geq xyz(x+y+z) \]
Which is obvious by AM-GM
Done!

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