\(\sum\dfrac{a^2-bc}{2a^2+b^2+c^2}=\dfrac{1}{2}\sum\dfrac{2a^2-2bc}{2a^2+b^2+c^2}=\dfrac{1}{2}\sum\dfrac{2a^2+b^2+c^2-\left(b+c\right)^2}{2a^2+b^2+c^2}\)
\(=\dfrac{1}{2}\sum\left(1-\dfrac{\left(b+c\right)^2}{2a^2+b^2+c^2}\right)\)
Nên ta chỉ cần chứng minh: \(\sum\dfrac{\left(b+c\right)^2}{2a^2+b^2+c^2}\le3\)
Thực vậy, ta có:
\(\sum\dfrac{\left(b+c\right)^2}{2a^2+b^2+c^2}=\sum\dfrac{\left(b+c\right)^2}{a^2+b^2+a^2+c^2}\le\sum\left(\dfrac{b^2}{a^2+b^2}+\dfrac{c^2}{a^2+c^2}\right)=3\)
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