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

Question

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

in progress 0
2 tháng 2021-10-02T14:59:51+00:00 2 Answers 5 views 0

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 $MinF = 30 \Leftrightarrow \left( {a;b;c} \right) = \left( {\dfrac{1}{3};\dfrac{1}{3};\dfrac{1}{3}} \right)$

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 $⇔a=b=c=\dfrac{1}{3}$