\(\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}+\dfrac{\left(c+a\right)^2}{5b^2+\left(c+a\right)^2}+\dfrac{\left(a+b\right)^2}{5c^2+\left(a+b\right)}\ge\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{-20a^2+10bc+5b^2+c^2}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{-20b^2+10ac+5c^2+5a^2}{9\left(5b^2+\left(c+a\right)^2\right)}+\dfrac{-20c^2+10ab+5a^2+5b^2}{9\left(5c^2+\left(a+b\right)\right)}\ge0\)
\(\Leftrightarrow\sum_{cyc}\dfrac{\left(c-a\right)\left(10a+5b+5c\right)-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\dfrac{-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{\left(a-b\right)\left(10b+5a+5c\right)}{9\left(5b^2+\left(a+c\right)^2\right)}\right)\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)\left(\dfrac{10b+5a+5c}{9\left(5b^2+\left(a+c\right)^2\right)}-\dfrac{10a+5b+5c}{9\left(5a^2+\left(b+c\right)^2\right)}\right)\right)\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)^2\dfrac{5\left(a^2+b^2-c^2+4ab\right)}{3\left(a^2+2ac+5b^2+c^2\right)\left(5a^2+b^2+2bc+c^2\right)}\right)\ge0\)
Dau "=" khi \(a=b=c\)
Bất đẳng thức cần chứng minh tương đương với
\(\dfrac{4}{3}-\sum\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}\le0\Leftrightarrow1-\sum\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}\le\dfrac{5}{3}\Leftrightarrow\sum\dfrac{5a^2}{5a^2+\left(b+c\right)^2}\le\dfrac{1}{3}\)\(\dfrac{9}{5a^2+\left(b+c\right)^2}=\dfrac{\left(1+2\right)^2}{a^2+b^2+c^2+2\left(2a^2+bc\right)}\le\dfrac{1}{a^2+b^2+c^2}+\dfrac{2}{2a^2+bc}\)
\(\Rightarrow\sum\dfrac{9a^2}{5a^2+\left(b+c\right)^2}\le\sum\dfrac{a^2}{a^2+b^2+c^2}+\sum\dfrac{2a^2}{2a^2+bc}=4-\sum\dfrac{bc}{2a^2+bc}\)Cần chứng minh \(\sum\dfrac{bc}{2a^2+bc}\ge1\). Ta có:
\(\sum\dfrac{bc}{2a^2+bc}\ge\dfrac{\left(\sum bc\right)^2}{\sum bc\left(2a^2+bc\right)}=1\)
Đẳng thức xảy ra khi \(a=b=c\) hoặc \(a=0;b=c\) và các hoán vị
