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\(VT=\sum\dfrac{a^2}{5a^2+b^2+c^2+2bc}=\sum\dfrac{a^2}{\left(2a^2+bc\right)+\left(2a^2+bc\right)+a^2+b^2+c^2}\)
\(\le\sum\dfrac{a^2}{9}\left(\dfrac{2}{2a^2+bc}+\dfrac{1}{a^2+b^2+c^2}\right)=\dfrac{1}{9}+\sum\dfrac{2a^2}{9\left(2a^2+bc\right)}\)
\(=\dfrac{4}{9}-\dfrac{1}{9}\left(\dfrac{bc}{2a^2+bc}+\dfrac{ac}{2b^2+ac}+\dfrac{ab}{2c^2+ab}\right)\)
\(\le\dfrac{4}{9}-\dfrac{1}{9}.\dfrac{\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2}=\dfrac{1}{3}\)
Dấu = xảy ra khi a=b=c
Ta có \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\)
\(=\sqrt{2a\left(a+b+c\right)+\dfrac{b^2-2bc+c^2}{2}}\)
\(=\sqrt{\dfrac{4a^2+b^2+c^2+4ab+4ac-2bc}{2}}\)
\(=\sqrt{\dfrac{\left(2a+b+c\right)^2-4bc}{2}}\)
\(\le\sqrt{\dfrac{\left(2a+b+c\right)^2}{2}}\)
\(=\dfrac{2a+b+c}{\sqrt{2}}\).
Vậy \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\le\dfrac{2a+b+c}{\sqrt{2}}\). Lập 2 BĐT tương tự rồi cộng vế, ta được \(VT\le\dfrac{2a+b+c+2b+c+a+2c+a+b}{\sqrt{2}}\)
\(=\dfrac{4\left(a+b+c\right)}{\sqrt{2}}\) \(=\dfrac{4.1011}{\sqrt{2}}\) \(=2022\sqrt{2}\)
ĐTXR \(\Leftrightarrow\) \(\left\{{}\begin{matrix}ab=0\\bc=0\\ca=0\\a+b+c=1011\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(1011;0;0\right)\) hoặc các hoán vị. Vậy ta có đpcm.
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
\(\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}+\dfrac{\left(c+a\right)^2}{5b^2+\left(c+a\right)^2}+\dfrac{\left(a+b\right)^2}{5c^2+\left(a+b\right)}\ge\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{-20a^2+10bc+5b^2+c^2}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{-20b^2+10ac+5c^2+5a^2}{9\left(5b^2+\left(c+a\right)^2\right)}+\dfrac{-20c^2+10ab+5a^2+5b^2}{9\left(5c^2+\left(a+b\right)\right)}\ge0\)
\(\Leftrightarrow\sum_{cyc}\dfrac{\left(c-a\right)\left(10a+5b+5c\right)-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\dfrac{-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{\left(a-b\right)\left(10b+5a+5c\right)}{9\left(5b^2+\left(a+c\right)^2\right)}\right)\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)\left(\dfrac{10b+5a+5c}{9\left(5b^2+\left(a+c\right)^2\right)}-\dfrac{10a+5b+5c}{9\left(5a^2+\left(b+c\right)^2\right)}\right)\right)\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)^2\dfrac{5\left(a^2+b^2-c^2+4ab\right)}{3\left(a^2+2ac+5b^2+c^2\right)\left(5a^2+b^2+2bc+c^2\right)}\right)\ge0\)
Dau "=" khi \(a=b=c\)
Nhung bo may thicc lam cach nay co duoc khong ?