Toán cho `:a;b;c>0` `a+b+c=1` tìm `minF=1/(a^2+b^2+c^2)+1/(ab)+1/(bc)+1/(ca)` 02/10/2021 By Julia cho `:a;b;c>0` `a+b+c=1` tìm `minF=1/(a^2+b^2+c^2)+1/(ab)+1/(bc)+1/(ca)`
Giải thích các bước giải: ĐK: $a,b,c>0$ Ta có: $\begin{array}{l}{\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {a – c} \right)^2} \ge 0,\forall a,b,c\\ \Leftrightarrow {a^2} + {b^2} + {c^2} \ge ab + bc + ac\\ \Leftrightarrow {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right) \ge 3\left( {ab + bc + ac} \right)\\ \Leftrightarrow {\left( {a + b + c} \right)^2} \ge 3\left( {ab + bc + ac} \right)\\ \Leftrightarrow 1 \ge 3\left( {ab + bc + ac} \right)\\ \Leftrightarrow ab + bc + ac \le \dfrac{1}{3}\end{array}$ Lại có: $\begin{array}{l}F = \dfrac{1}{{{a^2} + {b^2} + {c^2}}} + \dfrac{1}{{ab}} + \dfrac{1}{{bc}} + \dfrac{1}{{ac}}\\ = \left( {\dfrac{1}{{{a^2} + {b^2} + {c^2}}} + \dfrac{1}{{3ab}} + \dfrac{1}{{3bc}} + \dfrac{1}{{3ac}}} \right) + \left( {\dfrac{2}{{3ab}} + \dfrac{2}{{3bc}} + \dfrac{2}{{ac}}} \right)\\ = \left( {\dfrac{{{1^2}}}{{{a^2} + {b^2} + {c^2}}} + \dfrac{{{1^2}}}{{3ab}} + \dfrac{{{1^2}}}{{3bc}} + \dfrac{{{1^2}}}{{3ac}}} \right) + \dfrac{2}{3}\left( {\dfrac{{{1^2}}}{{ab}} + \dfrac{{{1^2}}}{{bc}} + \dfrac{{{1^2}}}{{ac}}} \right)\\ \ge \dfrac{{{{\left( {1 + 1 + 1 + 1} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + 3\left( {ab + bc + ac} \right)}} + \dfrac{2}{3}.\dfrac{{{{\left( {1 + 1 + 1} \right)}^2}}}{{ab + bc + ac}}\\\left( {BDT:Cauchy – Schwarz} \right)\\ = \dfrac{{16}}{{{{\left( {a + b + c} \right)}^2} + ab + bc + ac}} + \dfrac{2}{3}.\dfrac{9}{{ab + bc + ac}}\\ = \dfrac{{16}}{{1 + ab + bc + ac}} + \dfrac{6}{{ab + bc + ac}}\\ \ge \dfrac{{16}}{{1 + \dfrac{1}{3}}} + \dfrac{6}{{\dfrac{1}{3}}}\\ = 30\end{array}$ Dấu bằng xảy ra $\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}a = b = c\\a + b + c = 1\end{array} \right.\\ \Leftrightarrow a = b = c = \dfrac{1}{3}\end{array}$ Vậy $MinF = 30 \Leftrightarrow \left( {a;b;c} \right) = \left( {\dfrac{1}{3};\dfrac{1}{3};\dfrac{1}{3}} \right)$ Trả lời
Áp dụng bất đẳng thức Schwarz có: $\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}≥\dfrac{9}{ab+bc+ca}$ $⇒F≥\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}$ Áp dụng bất đẳng thức Schwarz ta có: $\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}≥\dfrac{9}{a^2+b^2+c^2+2(ab+bc+ca)}=\dfrac{9}{(a+b+c)^2}=9$ (do $a+b+c=1$ Lại có: $ab+bc+ca≤\dfrac{(a+b+c)^2}{3}=\dfrac{1}{3}$ (do $a+b+c=1$) $⇒\dfrac{7}{ab+bc+ca}≥7.3=21$ $⇒F≥\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}≥30$ Dấu $=$ xảy ra $⇔a=b=c=\dfrac{1}{3}$ Trả lời
Giải thích các bước giải:
ĐK: $a,b,c>0$
Ta có:
$\begin{array}{l}
{\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {a – c} \right)^2} \ge 0,\forall a,b,c\\
\Leftrightarrow {a^2} + {b^2} + {c^2} \ge ab + bc + ac\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right) \ge 3\left( {ab + bc + ac} \right)\\
\Leftrightarrow {\left( {a + b + c} \right)^2} \ge 3\left( {ab + bc + ac} \right)\\
\Leftrightarrow 1 \ge 3\left( {ab + bc + ac} \right)\\
\Leftrightarrow ab + bc + ac \le \dfrac{1}{3}
\end{array}$
Lại có:
$\begin{array}{l}
F = \dfrac{1}{{{a^2} + {b^2} + {c^2}}} + \dfrac{1}{{ab}} + \dfrac{1}{{bc}} + \dfrac{1}{{ac}}\\
= \left( {\dfrac{1}{{{a^2} + {b^2} + {c^2}}} + \dfrac{1}{{3ab}} + \dfrac{1}{{3bc}} + \dfrac{1}{{3ac}}} \right) + \left( {\dfrac{2}{{3ab}} + \dfrac{2}{{3bc}} + \dfrac{2}{{ac}}} \right)\\
= \left( {\dfrac{{{1^2}}}{{{a^2} + {b^2} + {c^2}}} + \dfrac{{{1^2}}}{{3ab}} + \dfrac{{{1^2}}}{{3bc}} + \dfrac{{{1^2}}}{{3ac}}} \right) + \dfrac{2}{3}\left( {\dfrac{{{1^2}}}{{ab}} + \dfrac{{{1^2}}}{{bc}} + \dfrac{{{1^2}}}{{ac}}} \right)\\
\ge \dfrac{{{{\left( {1 + 1 + 1 + 1} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + 3\left( {ab + bc + ac} \right)}} + \dfrac{2}{3}.\dfrac{{{{\left( {1 + 1 + 1} \right)}^2}}}{{ab + bc + ac}}\\
\left( {BDT:Cauchy – Schwarz} \right)\\
= \dfrac{{16}}{{{{\left( {a + b + c} \right)}^2} + ab + bc + ac}} + \dfrac{2}{3}.\dfrac{9}{{ab + bc + ac}}\\
= \dfrac{{16}}{{1 + ab + bc + ac}} + \dfrac{6}{{ab + bc + ac}}\\
\ge \dfrac{{16}}{{1 + \dfrac{1}{3}}} + \dfrac{6}{{\dfrac{1}{3}}}\\
= 30
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
a = b = c\\
a + b + c = 1
\end{array} \right.\\
\Leftrightarrow a = b = c = \dfrac{1}{3}
\end{array}$
Vậy $MinF = 30 \Leftrightarrow \left( {a;b;c} \right) = \left( {\dfrac{1}{3};\dfrac{1}{3};\dfrac{1}{3}} \right)$
Áp dụng bất đẳng thức Schwarz có:
$\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}≥\dfrac{9}{ab+bc+ca}$
$⇒F≥\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}$
Áp dụng bất đẳng thức Schwarz ta có:
$\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}≥\dfrac{9}{a^2+b^2+c^2+2(ab+bc+ca)}=\dfrac{9}{(a+b+c)^2}=9$ (do $a+b+c=1$
Lại có: $ab+bc+ca≤\dfrac{(a+b+c)^2}{3}=\dfrac{1}{3}$ (do $a+b+c=1$)
$⇒\dfrac{7}{ab+bc+ca}≥7.3=21$
$⇒F≥\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}≥30$
Dấu $=$ xảy ra $⇔a=b=c=\dfrac{1}{3}$