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pt co nghiem nguyen khi \(\Delta \) la SCP
\(\Delta=\left(-\left(m+4\right)\right)^2-4\cdot2m\)
\(=m^2+8m+16-8m=m^2+16>0\forall m\)
Dat \(m^2+16=a^2\left(a\in Z\right)\)
\(\Leftrightarrow\left(a-m\right)\left(a+m\right)=16\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m=0\\a=\pm4\end{matrix}\right.\\\left\{{}\begin{matrix}m=\pm3\\a=\pm5\end{matrix}\right.\end{matrix}\right.\)
SOS get it <(")
\(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)->\left(a;;bc\right)\text{for}\left(a;b;c>0\text{and}a^2+b^2+c^2=1\right)\)
\(\text{Khido}P=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)
\(\text{Ta se cm}\sum_{cyc}\frac{a}{b^2+c^2}\ge\frac{3\sqrt{3}}{2}\)\(\text{Viet lai BDT can chung minh}\)
\(\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3\sqrt{3}}{2\sqrt{x^2+y^2+z^2}}\)
\(\text{Chuan hoa}a^2+b^2+c^2=3\text{ta can cm:}\)
\(\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{3-a^2}+\frac{b}{3-b^2}+\frac{c}{3-c^2}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{3-a^2}-\frac{1}{2}+\frac{b}{3-b^2}-\frac{1}{2}+\frac{c}{3-c^2}-\frac{1}{2}\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\frac{a}{3-a^2}-\frac{1}{2}-\frac{1}{2}\left(x^2-1\right)\right)\ge0\)
\(\Leftrightarrow\frac{a\left(a+2\right)\left(a-1\right)^2}{3-a^2}+\frac{b\left(b+2\right)\left(b-1\right)^2}{3-b^2}+\frac{c\left(c+2\right)\left(c-1\right)^2}{3-c^2}\ge0\)