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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 \(min_P=7\)
\(A=\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\)
\(=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\dfrac{34}{ab}+\dfrac{17}{8}ab-\dfrac{1}{8}ab\)
\(\ge2.\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{34}{ab}.\dfrac{17}{8}ab}-\dfrac{1}{8}.\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow A\ge2.\dfrac{4}{\left(a+b\right)^2}+2.\dfrac{17}{2}-\dfrac{1}{8}.\dfrac{4^2}{4}\ge2.\dfrac{4}{4^2}+17-\dfrac{1}{2}\)
\(\Leftrightarrow A\ge\dfrac{1}{2}+17-\dfrac{1}{2}=17\)
Dấu "=" <=> a = b = 2
\(2ab+6bc+2ac=7abc\\ \Rightarrow\dfrac{2}{c}+\dfrac{6}{a}+\dfrac{2}{b}=7\\ \)
Đặt x=1/a ; y=1/b ; z=1/c
\(\Rightarrow6x+2y+2z=7\)
\(H=\dfrac{4}{\dfrac{1}{b}+\dfrac{2}{a}}+\dfrac{9}{\dfrac{1}{c}+\dfrac{4}{a}}+\dfrac{4}{\dfrac{1}{c}+\dfrac{1}{b}}\\ =\dfrac{4}{2x+y}+\dfrac{9}{z+4x}+\dfrac{4}{y+z}\)
BĐT Cô si ;
\(\left(\dfrac{4}{2x+y}+\left(2x+y\right)\right)+\left(\dfrac{9}{4x+z}+\left(4x+z\right)\right)+\left(\dfrac{4}{y+z}+\left(y+z\right)\right)\\ \ge2\sqrt{4}+2\sqrt{9}+2\sqrt{4}=14\\ \Rightarrow C+7\ge14\\ \Rightarrow C\ge7\)
Min C=7 khi a=2;b=c=1