De dung la:
\(\Sigma_{cyc}\frac{1}{1+a^2+b^2}\le\frac{9}{5}\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a^2+b^2}{1+a^2+b^2}\ge\frac{6}{5}\)
\(VT\ge\frac{\left(\Sigma_{cyc}\sqrt{a^2+b^2}\right)^2}{2\Sigma_{cyc}a^2+3}\left(M\right)\)
Consider:
\(VT_M\ge\frac{6}{5}\)
\(5\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge\Sigma_{cyc}a^2+9\)
Consider:
\(5\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge5\Sigma_{cyc}a^2+5\Sigma_{cyc}ab=5\Sigma_{cyc}a^2+5\)
Gio can cung minh:
\(5\Sigma_{cyc}a^2+5\ge\Sigma_{cyc}a^2+9\)
\(\Leftrightarrow\Sigma_{cyc}a^2\ge1\)
Ta lai co:
\(\Sigma_{cyc}a^2\ge\Sigma_{cyc}ab=1\)
Dau '=' xay ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
