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\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
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 \(a=b=c\)
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
\(ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\) (do a,b,c là các số dương)
Áp dụng BĐT Bunhiacopxki dạng phân thức:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{6^2}{a+2b+3c}\)
\(\Rightarrow\dfrac{36}{a+2b+3c}\le\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\left(1\right)\)
Tương tự: \(\left\{{}\begin{matrix}\dfrac{36}{b+2c+3a}\le\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\left(2\right)\\\dfrac{36}{c+2a+3b}\le\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\left(3\right)\end{matrix}\right.\)
Lấy (1) + (2) + (3) ta được:
\(36F\le6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=6.3=18\)
\(\Rightarrow F\le\dfrac{1}{2}\)
MaxF=1/2 khi \(a=b=c=1\)
Bổ đề :\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)
Áp dụng bất đẳng thức Cô-si ta có:
\(x+y+z\ge3\sqrt[3]{xyz};\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge3\sqrt[3]{\dfrac{1}{x}.\dfrac{1}{y}.\dfrac{1}{z}}\)
\(\Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\dfrac{1}{x}\dfrac{1}{y}\dfrac{1}{z}}=9\)
Dấu "=" xảy ra ⇔ x=y=z
Ta có:\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{9}.\dfrac{9}{a+3b+2c}\le\dfrac{ab}{9}.\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)\)
Tương tự ta có:\(\dfrac{bc}{b+3c+2a}\le\dfrac{bc}{9}\left(\dfrac{1}{b+a}+\dfrac{1}{c+a}+\dfrac{1}{2c}\right)\)
\(\dfrac{ca}{c+3a+2b}\le\dfrac{ca}{9}.\left(\dfrac{1}{c+b}+\dfrac{1}{a+b}+\dfrac{1}{2a}\right)\)
Cộng vế với vế ta có:
\(A\le\dfrac{1}{9}.\left(\dfrac{ab+bc}{a+c}+\dfrac{cb+ac}{a+b}+\dfrac{ca+ab}{b+c}+\dfrac{a+b+c}{2}\right)\)
\(=\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}.\left(6+\dfrac{6}{3}\right)=1\)
Dấu "=" xảy ra ⇔ a=b=c=2
Vậy Max A=1⇔ a=b=c=2
\(BDT\Leftrightarrow2a^4b+2b^4c+2c^4a+3ab^4+3bc^4+3ca^4\ge5a^2b^2c+5a^2bc^2+5ab^2c^2\)
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 \(a=b=c\)
\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)
Tương tự và cộng lại;
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng bđt \(\dfrac{9}{a+b+c}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Khi đó \(\dfrac{9.ab}{a+3b+2c}=ab.\dfrac{9}{\left(a+c\right)+\left(c+b\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{c+b}+\dfrac{a}{2}\)
Tương tự và cộng theo vế suy ra \(9A\le\dfrac{3\left(a+b+c\right)}{2}=9< =>A\le1\)
Dấu "=" xảy ra khi và chỉ khi a = b = c = 2
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