cho `:a;b;c>0` `a+b+c=1` tìm `minF=1/(a^2+b^2+c^2)+1/(ab)+1/(bc)+1/(ca)`

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1. 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

2. 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