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C/m BĐT : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)
Áp dụng BĐT Sơ-vác-sơ:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}\ge\dfrac{9}{x+y+z}\)
Ta có: \(9\dfrac{ab}{a+3b+2c}=\dfrac{9ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\left(1\right)\)
CM tương tự
\(\dfrac{9bc}{b+3c+2a}\le\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{b}{2}\left(2\right)\)
\(\dfrac{9ca}{c+3a+2b}\le\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\left(3\right)\)
Cộng vế (1), (2), (3) => đpcm
a.
\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)
2.
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)
Quay lại câu a
Áp dụng bất đẳng thức Cauchy-Schwarz:\(\left\{{}\begin{matrix}\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}\ge\dfrac{\left(1+1\right)^2}{a+2b+c+c+3a}=\dfrac{4}{4a+2b+2c}=\dfrac{2}{c+2a+b}\\\dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}\ge\dfrac{\left(1+1\right)^2}{b+2c+a+a+3b}=\dfrac{4}{4b+2c+2a}=\dfrac{2}{a+2b+c}\\\dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}\ge\dfrac{\left(1+1\right)^2}{c+2a+b+b+3c}=\dfrac{4}{4c+2a+2b}=\dfrac{2}{b+2c+a}\end{matrix}\right.\)
Cộng theo vế ta có:
\(\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}+\dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}+\dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}\ge\dfrac{2}{c+2a+b}+\dfrac{2}{a+2b+c}+\dfrac{2}{b+2c+a}\)
Hay \(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\le\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\left(đpcm\right)\)
Áp dụng BĐT Cô si dạng Engel ; ta có :
\(\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}\ge\dfrac{\left(1+1\right)^2}{\left(a+2b+c\right)+\left(c+3a\right)}=\dfrac{4}{4a+2b+2c}=\dfrac{2}{2a+b+c}\\ \dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}\ge\dfrac{\left(1+1\right)^2}{\left(b+2c+a\right)+\left(a+3b\right)}=\dfrac{4}{4b+2c+2a}=\dfrac{2}{2b+c+a}\\ \dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}\ge\dfrac{\left(1+1\right)^2}{\left(c+2a+b\right)+\left(b+3c\right)}=\dfrac{4}{4c+2a+2b}=\dfrac{2}{2c+a+b}\)
\(\Rightarrow\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}+\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{2}{a+2b+c}+\dfrac{2}{b+2c+a}+\dfrac{2}{c+2a+b}\\ \Rightarrow\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\)
\(BDT\Leftrightarrow2a^4b+2b^4c+2c^4a+3ab^4+3bc^4+3ca^4\ge5a^2b^2c+5a^2bc^2+5ab^2c^2\)
Ta chứng minh được \(ab^4+bc^4+ca^4\ge a^2b^2c+a^2bc^2+ab^2c^2\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)
\(VT=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=VP\)
Vậy ta cần chứng minh \(2a^4b+2b^4c+2c^4a+2ab^4+2bc^4+2ca^4\ge4a^2b^2c+4a^2bc^2+4ab^2c^2\)
\(\Leftrightarrow\sum_{cyc}\left(2c^3+bc^2-b^2c+ac^2-a^2c+3ab^2+3a^2b\right)\left(a-b\right)^2\ge0\)
Dấu "=" xảy ra khi \(a=b=c\)
b) \(\dfrac{1}{3a+2b+c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{36}\left(\dfrac{3}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)
Tương tự cho 2 cái kia rồi cộng lại
\(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}.16=\dfrac{8}{3}\)
Đẳng thức xảy ra \(\Leftrightarrow\) ... \(\Leftrightarrow a=b=c=\dfrac{3}{16}\)
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=\dfrac{1}{a}+\dfrac{4}{2b}+\dfrac{9}{3c}\ge\dfrac{\left(1+2+3\right)^2}{a+2b+3c}=\dfrac{36}{a+2b+3c}\)
\(\dfrac{2}{a}+\dfrac{3}{b}+\dfrac{1}{c}=\dfrac{4}{2a}+\dfrac{9}{3b}+\dfrac{1}{c}\ge\dfrac{\left(2+3+1\right)^2}{2a+3b+c}=\dfrac{36}{2a+3b+c}\)
\(\dfrac{3}{a}+\dfrac{1}{b}+\dfrac{2}{c}=\dfrac{9}{3a}+\dfrac{1}{b}+\dfrac{4}{2c}\ge\dfrac{\left(3+1+2\right)^2}{3a+b+2c}=\dfrac{36}{3a+2b+c}\)
Cộng theo vế: \(6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge36F\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge6F\)
Mặt khác: \(ab+bc+ac=3abc\Leftrightarrow\dfrac{ab+bc+ac}{abc}=3\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
\(\Rightarrow18\ge36F\Leftrightarrow F\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Tương tự:
\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{c+a}+\dfrac{b}{2}\right)\)
\(\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ca+ab}{b+c}+\dfrac{bc+ab}{c+a}+\dfrac{a+b+c}{2}\right)=\dfrac{a+b+c}{6}\)
Dấu "=" xảy ra khi \(a=b=c\)