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\(\frac{4c}{4c+57}\ge\frac{1}{1+a}+\frac{35}{35+2b}\ge2\sqrt{\frac{35}{\left(1+a\right)\left(35+2b\right)}}\)
\(\frac{a}{1+a}\ge\frac{57}{4c+57}+\frac{35}{35+2b}\ge2\sqrt{\frac{35\cdot57}{\left(4c+57\right)\left(35+2b\right)}}\)
\(\frac{2b}{35+2b}\ge\frac{57}{4c+57}+\frac{1}{1+a}\ge2\sqrt{\frac{57}{\left(4c+57\right)\left(1+a\right)}}\)
\(\Rightarrow8abc\ge8\cdot1995\Rightarrow abc\ge1995\)
Vậy giá trị nhỏ nhất của abc là 1995
\(P=\dfrac{4ab}{a+2b}+\dfrac{9ca}{a+4c}+\dfrac{4bc}{b+c}\)
\(P=\dfrac{4abc}{ac+2bc}+\dfrac{9abc}{ab+4bc}+\dfrac{4abc}{ab+ac}\)
\(P=abc\left(\dfrac{4}{ac+2bc}+\dfrac{9}{ab+4bc}+\dfrac{4}{ab+ac}\right)\)
\(P\ge abc.\dfrac{\left(2+3+2\right)^2}{ac+2bc+ab+4bc+ab+ac}\)
\(P\ge abc.\dfrac{49}{2ab+6bc+2ca}\)
\(P\ge abc.\dfrac{49}{7abc}\) (vì \(2ab+6bc+2ca=7abc\))
\(P\ge7\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{ac+2bc}=\dfrac{3}{ab+4bc}=\dfrac{2}{ab+ac}\\2ab+6bc+2ca=7abc\end{matrix}\right.\)
\(\dfrac{2}{ac+2bc}=\dfrac{2}{ab+ac}\) \(\Leftrightarrow2b=a\)
Có \(\dfrac{3}{ab+4bc}=\dfrac{2}{ab+ac}\)
\(\Leftrightarrow\dfrac{3}{2b^2+4bc}=\dfrac{2}{2b^2+2bc}\)
\(\Leftrightarrow3b^2+3bc=2b^2+4bc\)
\(\Leftrightarrow b^2=bc\Leftrightarrow b=c\)
\(\Rightarrow a=2b=2c\)
Lại có \(2ab+6bc+2ca=7abc\) \(\Rightarrow4b^2+6b^2+4b^2=14b^3\)
\(\Leftrightarrow b=1\)
\(\Leftrightarrow\left(a,b,c\right)=\left(2,1,1\right)\)
Vậy \(min_P=7\)
Ta có:
\(M=\dfrac{4ab}{a+2b}+\dfrac{9ac}{a+4c}+\dfrac{4bc}{b+c}\)
\(=\dfrac{4}{\dfrac{1}{b}+\dfrac{2}{a}}+\dfrac{9}{\dfrac{1}{c}+\dfrac{4}{a}}+\dfrac{4}{\dfrac{1}{c}+\dfrac{1}{b}}\)
\(\ge\dfrac{\left(2+3+2\right)^2}{\dfrac{1}{b}+\dfrac{2}{a}+\dfrac{1}{c}+\dfrac{4}{a}+\dfrac{1}{c}+\dfrac{1}{b}}=\dfrac{49}{\dfrac{2}{b}+\dfrac{6}{a}+\dfrac{2}{c}}=\dfrac{49}{\dfrac{2ab+6bc+2ac}{abc}}=\dfrac{49}{7}=7\)
Vậy GTNN là M = 7 khi \(\left(a,b,c\right)=\left(2,1,1\right)\)
\(\Leftrightarrow\frac{1}{1+a}+\frac{a}{1+a}+\frac{2b}{21+2b}+\frac{21}{21+2b}\le\frac{4c}{4c+27}+\frac{a}{1+a}+\frac{2b}{21+2b}\)
\(\Leftrightarrow2\le\frac{1}{1+\frac{1}{a}}+\frac{1}{1+\frac{21}{2b}}+\frac{1}{1+\frac{27}{4c}}\)
Đặt \(\left(\frac{1}{a};\frac{21}{2b};\frac{27}{4c}\right)=\left(x;y;z\right)\)
\(\Leftrightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge2\)
\(\Leftrightarrow\frac{1}{1+x}\ge1-\frac{1}{1+y}+1-\frac{1}{1+z}=\frac{y}{1+y}+\frac{z}{1+z}\ge2\sqrt{\frac{yz}{\left(1+y\right)\left(1+z\right)}}\)
Tương tự: \(\frac{1}{1+y}\ge2\sqrt{\frac{zx}{\left(1+z\right)\left(1+x\right)}}\) ; \(\frac{1}{1+z}\ge2\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Nhân vế với vế: \(1\ge8xyz\Rightarrow xyz\le\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{a}.\frac{21}{2b}.\frac{27}{4c}\le\frac{1}{8}\Leftrightarrow abc\ge567\)
Dấu "=" xảy ra khi \(\frac{1}{a}=\frac{21}{2b}=\frac{27}{4c}=\frac{1}{2}\Rightarrow\left(a;b;c\right)=\left(2;21;\frac{27}{2}\right)\)