Cách này được không ạ?
\(P=\left(a^2+b^2+c^2\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(\Rightarrow P\ge\left[\frac{\left(a+b\right)^2}{2}+c^2\right]\left[\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)^2+\frac{1}{c^2}\right]\)
\(\Rightarrow P\ge\left[\frac{\left(a+b\right)^2}{2}+c^2\right]\left[\frac{1}{2}\left(\frac{4}{a+b}\right)^2+\frac{1}{c^2}\right]\)
\(\Rightarrow P\ge\left[\frac{\left(a+b\right)^2}{2}+c^2\right]\left[\frac{8}{\left(a+b\right)^2}+\frac{1}{c^2}\right]\)
Đặt: \(\left\{{}\begin{matrix}a+b=x\\c=y\end{matrix}\right.\Rightarrow x\le y\)
\(\Rightarrow P\ge\left(\frac{x^2}{2}+y^2\right)\left(\frac{8}{x^2}+\frac{1}{y^2}\right)\)
\(\Rightarrow P\ge4+1+\frac{8y^2}{x^2}+\frac{x^2}{2y^2}\)
\(\Rightarrow P\ge5+\frac{15y^2}{2x^2}+\frac{1}{2}\left(\frac{y^2}{x^2}+\frac{x^2}{y^2}\right)\)
\(\Rightarrow P\ge5+\frac{15}{2}+\frac{1}{2}.2\sqrt{\frac{y^2}{x^2}.\frac{x^2}{y^2}}\left(y\ge x\right)\)
\(\Rightarrow P\ge5+\frac{15}{2}+\frac{1}{2}.2\)
\(\Rightarrow P\ge\frac{27}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{c}{2}\)
Ta có:
\( P = \left( {\dfrac{{{a^2}}}{{{b^2}}} + \dfrac{{{b^2}}}{{{a^2}}}} \right) + {c^2}\left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} \right) + \dfrac{1}{{{c^2}}}\left( {{a^3} + {b^3}} \right) + 3\\ P = \left( {\dfrac{{{a^2}}}{{{b^2}}} + \dfrac{{{b^2}}}{{{a^2}}}} \right) + \left( {\dfrac{{{c^2}}}{{16{a^2}}} + \dfrac{{{a^3}}}{{{c^2}}}} \right) + \left( {\dfrac{{{c^2}}}{{16{b^2}}} + \dfrac{{{b^2}}}{{{c^2}}}} \right) + \dfrac{{15{c^2}}}{{16}}\left( {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} \right) \)
Tự tiếp nhé!
Ta dự đoán: \(Min_P\Leftrightarrow\left\{{}\begin{matrix}a+b=c\\a=b\end{matrix}\right.\) \(\Leftrightarrow a=b=\frac{c}{2}\)
Từ trên ta suy ra được:
Ta có: \(P=\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)+c^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{1}{c^2}\left(a^2+b^2\right)+3\)
\(P=\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)+\left(\frac{c^2}{16a^2}+\frac{a^2}{c^2}\right)+\left(\frac{c^2}{16b^2}+\frac{b^2}{c^2}\right)+\frac{15c^2}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)
\(\Rightarrow Min_P=\frac{27}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{c}{2}\)
