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Ta có:
\(2ab+6bc+2ca=7abc\)
Chia cả hai vế của phương trình trên cho \(abc>0\), ta được:
\(\frac{6}{a}+\frac{2}{b}+\frac{2}{c}=7\)
Đặt \(x=\frac{2}{a};\) \(y=\frac{1}{b};\) và \(z=\frac{1}{c}\) \(\Rightarrow\) \(\hept{\begin{cases}x,y,z\in Z_+\\3x+2y+2z=7\end{cases}}\)
Khi đó, ta biểu diễn biểu thức \(C\) dưới dạng ba biến \(x,y,z\) như sau:
\(C=\frac{4ab}{a+2b}+\frac{9ca}{a+4c}+\frac{4bc}{b+c}=\frac{4}{x+y}+\frac{9}{z+2x}+\frac{4}{y+z}\)
nên \(C=\left[\frac{4}{x+y}+\left(x+y\right)\right]+\left[\frac{9}{z+2x}+\left(z+2x\right)\right]+\left[\frac{4}{y+z}+\left(y+z\right)\right]-\left(3x+2y+2z\right)\)
Áp dụng bất đẳng thức \(AM-GM\) cho từng bộ số trong ngoặc luôn dương, ta có:
\(C\ge4+6+4-7=7\) (do \(3x+2y+2z=7\) )
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\frac{4}{x+y}=x+y\\\frac{9}{z+2x}=z+2x\\\frac{4}{y+z}=y+z\end{cases}}\) \(\Leftrightarrow\) \(x=y=z=1\)
Do đó, \(a=2;\) và \(y=z=1\)
Vậy, \(GTNN\) của \(C\) đạt được là \(7\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=2\\y=z=1\end{cases}}\)
\(C=\frac{4ab}{a+2b}+\frac{9ac}{4c+a}+\frac{4bc}{b+c}=\frac{4abc}{ac+2bc}+\frac{9abc}{4bc+ab}+\frac{4abc}{ab+ac}\)
\(\ge\frac{\left(2\sqrt{abc}+3\sqrt{abc}+2\sqrt{abc}\right)^2}{ac+2bc+4bc+ab+ab+ac}=\frac{49abc}{2ac+6bc+2ab}=7\)
Xin bổ sung cách sau, bn có thể tham khảo thêm
:\(GT\Leftrightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)
Đặt \(\hept{\begin{cases}\frac{1}{c}=x\\\frac{1}{b}=y\\\frac{3}{a}=z\end{cases}}\) Ta có: \(2\left(x+y+z\right)=7\)
Suy ra \(C=\frac{4}{4y+\frac{2z}{3}}+\frac{9}{x+\frac{4z}{3}}+\frac{4}{x+y}\ge\frac{\left(2+3+2\right)^2}{2\left(x+y+z\right)}=7\) (Bdt Cauchy-Schwarz)
Dấu = khi \(\hept{\begin{cases}a=2\\b=c=1\end{cases}}\)
\(\frac{4c}{4c+57}\ge\frac{1}{1+a}+\frac{35}{35+2b}\ge2\sqrt{\frac{35}{\left(1+a\right)\left(35+2b\right)}}\)
\(\frac{a}{1+a}\ge\frac{57}{4c+57}+\frac{35}{35+2b}\ge2\sqrt{\frac{35\cdot57}{\left(4c+57\right)\left(35+2b\right)}}\)
\(\frac{2b}{35+2b}\ge\frac{57}{4c+57}+\frac{1}{1+a}\ge2\sqrt{\frac{57}{\left(4c+57\right)\left(1+a\right)}}\)
\(\Rightarrow8abc\ge8\cdot1995\Rightarrow abc\ge1995\)
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 \(min_P=7\)
2ab + 6bc + 2ac = 7abc => \(\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)
đặt \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\) => 6x + 2y + 2z = 7; x; y; z > 0
Khi đó, C = \(\frac{4}{\frac{1}{b}+\frac{2}{a}}+\frac{9}{\frac{1}{c}+\frac{4}{a}}+\frac{4}{\frac{1}{c}+\frac{1}{b}}=\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)
AD BĐT Cauchy ta có:
\(\left(\frac{4}{2x+y}+\left(2x+y\right)\right)+\left(\frac{9}{4x+z}+\left(4x+z\right)\right)+\left(\frac{4}{y+z}+\left(y+z\right)\right)\)
\(\ge2\sqrt{4}+2.\sqrt{9}+2.\sqrt{4}=14\)
=> \(\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)+ 7 > 14 => C > 7
Dấu "=" xảy ra <=> a = 2; b = 1; c = 1
Vậy Min C = 7
2ab+6bc+2ac=7abc =>
Đặt => 6x + 2y + 2z = 7; x; y; z > 0
Khi đó C=
TA CÓ:
Dấu “=” xảy raóa=2;b=1;c=1
Vậy c=7
Xong rồi đó bạn hứa cho mik nha
Bài 1: Theo đề : \(2ab+6bc+2ac=7abc\) \(;a,b,c>0\)
Chia cả 2 vế cho \(abc>0\Rightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)
Đặt: \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}x,y,z>0\\2z+6x+2y=7\end{cases}}\)
Khi đó: \(M=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}=\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)
\(\Rightarrow M=\frac{4}{2x+y}+2x+y+\frac{9}{4x+z}+4x+z+\frac{4}{y+z}+y+z-\left(2x+y+4x+z+y+z\right)\)
\(=\left(\frac{2}{\sqrt{x+2y}}-\sqrt{x+2y}\right)^2+\left(\frac{3}{\sqrt{4x+z}}-\sqrt{4x+z}\right)^2+\left(\frac{2}{\sqrt{y+z}}-\sqrt{y+z}\right)^2+17\ge17\)
Khi: \(\hept{\begin{cases}x=\frac{1}{2}\\y=z=1\end{cases}}\Rightarrow M=17\)
\(Min_M=17\Leftrightarrow a=2;b=1;c=1\)
ミ★๖ۣۜBăηɠ ๖ۣۜBăηɠ ★彡 chém bài khó nhất rồi nên em xin mạn phép chém bài dễ ạ.
2/\(VT=\Sigma_{cyc}\frac{\left(x+y+z\right)^2-x^2}{x\left(x+y+z\right)+yz}=\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}\)
\(\ge\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\frac{\left(2x+y+z\right)^2}{4}}=\Sigma_{cyc}\frac{4\left(y+z\right)}{2x+y+z}=\Sigma_{cyc}\frac{2\left(y+z-2x\right)}{2x+y+z}+6\)
\(=\Sigma_{cyc}\left(\frac{2\left(x+y+z\right)\left(y+z-2x\right)}{2x+y+z}-\frac{3}{2}\left(y+z-2x\right)\right)+6\)
\(=\Sigma_{cyc}\frac{\left(y+z-2x\right)^2}{2\left(2x+y+z\right)}+6\ge6\)
\(M=\frac{\left(a+1\right)^2+2a}{a\left(a+1\right)}+\frac{\left(b+1\right)^2+2b}{b\left(b+1\right)}+\frac{\left(c+1\right)^2+2c}{c\left(c+1\right)}\)
\(M=\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+1}{c}+2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)
\(M=3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)
\(M\ge3+\frac{9}{a+b+c}+2\left(\frac{9}{a+b+c+3}\right)\ge3+3+3=9\)
Dấu "=" xảy ra khi a=b=c=1
\(2ab+5bc+6ca=6abc\Rightarrow\frac{5}{a}+\frac{6}{b}+\frac{2}{c}=6\)
\(P=\frac{1}{\frac{1}{a}+\frac{2}{b}}+\frac{4}{\frac{4}{b}+\frac{1}{c}}+\frac{9}{\frac{1}{c}+\frac{4}{a}}\ge\frac{\left(1+2+3\right)^2}{\frac{5}{a}+\frac{6}{b}+\frac{2}{c}}=6\)
\(P_{min}=6\) khi \(\left\{{}\begin{matrix}\frac{1}{a}+\frac{2}{b}=\frac{2}{b}+\frac{1}{2c}=\frac{1}{3c}+\frac{4}{3a}\\\frac{5}{a}+\frac{6}{b}+\frac{2}{c}=6\end{matrix}\right.\)
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