Cho các số thực dương a, b, c thoả mãn a + b + c = 3.
CMR:$\frac{3a-{{a}^{2}}}{4-bc}+\frac{3b-{{b}^{2}}}{4-ca}+\frac{3c-{{c}^{2}}}{4-ab}\ge 2abc\left( * \right)$
Giải:
Ta có $\ \left( * \right)\Leftrightarrow \frac{3-a}{4bc-{{\left( bc \right)}^{2}}}+\frac{3-b}{4ca-{{\left( ca \right)}^{2}}}+\frac{3-c}{4ab-{{\left( ab \right)}^{2}}}\ge 2.$
$\ \Leftrightarrow \frac{b+c}{4bc-{{\left( bc \right)}^{2}}}+\frac{c+a}{4ca-{{\left( ca \right)}^{2}}}+\frac{a+b}{4ab-{{\left( ab \right)}^{2}}}\ge 2\left( 1 \right).$
Áp dụng BĐT Côsi ta có: $\ \left\{ \begin{align}
& b+c\ge 2\sqrt{bc} \\
& c+a\ge 2\sqrt{ca} \\
& a+b\ge 2\sqrt{ab} \\
\end{align} \right.\Rightarrow \left( 1 \right)\Leftrightarrow \frac{\sqrt{bc}}{4bc-{{\left( bc \right)}^{2}}}+\frac{\sqrt{ca}}{4ca-{{\left( ca \right)}^{2}}}+\frac{\sqrt{ab}}{4ab-{{\left( ab \right)}^{2}}}\ge 1.$
$\ \Leftrightarrow \frac{1}{4\sqrt{ab}-ab\sqrt{ab}}+\frac{1}{4\sqrt{bc}-bc\sqrt{bc}}+\frac{1}{4\sqrt{ca}-ca\sqrt{ca}}\ge 1\left( 2 \right).$
Đặt: $\ \left\{ \begin{array}{l}
x = \sqrt {ab} \\
y = \sqrt {bc} \\
z = \sqrt {ca}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + y + z = \sqrt {ab} + \sqrt {bc} + \sqrt {ca} \le a + b + c = 3\\
x,y,z > 0
\end{array} \right.$
$\ \Rightarrow \left( 2 \right)\Leftrightarrow \frac{1}{4x-{{x}^{3}}}+\frac{1}{4y-{{y}^{3}}}+\frac{1}{4z-{{z}^{3}}}\ge 1\left( 3 \right).$
Ta luôn có: $\ {{\left( x-1 \right)}^{2}}\left( x+2 \right)\ge 0\Leftrightarrow {{x}^{3}}-3x+2\ge 0\Leftrightarrow 4x-{{x}^{3}}\le x+2\Rightarrow \frac{1}{4x-{{x}^{3}}}\ge \frac{1}{x+2}.$
$\ \Rightarrow \frac{1}{4x-{{x}^{3}}}+\frac{1}{4y-{{y}^{3}}}+\frac{1}{4z-{{z}^{3}}}\ge \frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\ge \frac{9}{\left( x+y+z \right)+6}\ge \frac{9}{3+6}=1\Rightarrow $ ĐPCM
Dấu = xảy ra khi và chỉ khi $\left\{ \begin{align}
& a=b=c=1 \\
& a+b+c=3 \\
& ab=bc=ca=1 \\
\end{align} \right.\Leftrightarrow a=b=c=1$ .
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