Giúp mình nha!
Cho a, b, c >0 thỏa mãn:
\(\dfrac{1}{1+a}+\dfrac{21}{21+2b}\le\dfrac{4c}{96+4c}\)
Tìm giá trị nhỏ nhất của \(T=a.b.c\)
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TN
24 tháng 6 2016
\(\frac{4c}{4c+57}\ge\frac{1}{1+a}+\frac{35}{35+2b}\ge2\sqrt{\frac{35}{\left(1+a\right)\left(35+2b\right)}}\)
\(\frac{a}{1+a}\ge\frac{57}{4c+57}+\frac{35}{35+2b}\ge2\sqrt{\frac{35\cdot57}{\left(4c+57\right)\left(35+2b\right)}}\)
\(\frac{2b}{35+2b}\ge\frac{57}{4c+57}+\frac{1}{1+a}\ge2\sqrt{\frac{57}{\left(4c+57\right)\left(1+a\right)}}\)
\(\Rightarrow8abc\ge8\cdot1995\Rightarrow abc\ge1995\)
Vậy giá trị nhỏ nhất của abc là 1995
4 tháng 11 2023
\(P=\dfrac{4ab}{a+2b}+\dfrac{9ca}{a+4c}+\dfrac{4bc}{b+c}\)
\(P=\dfrac{4abc}{ac+2bc}+\dfrac{9abc}{ab+4bc}+\dfrac{4abc}{ab+ac}\)
\(P=abc\left(\dfrac{4}{ac+2bc}+\dfrac{9}{ab+4bc}+\dfrac{4}{ab+ac}\right)\)
\(P\ge abc.\dfrac{\left(2+3+2\right)^2}{ac+2bc+ab+4bc+ab+ac}\)
\(P\ge abc.\dfrac{49}{2ab+6bc+2ca}\)
\(P\ge abc.\dfrac{49}{7abc}\) (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 \(min_P=7\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 3 2023
\(2ab+6bc+2ac=7abc\Rightarrow\dfrac{6}{a}+\dfrac{2}{b}+\dfrac{2}{c}=7\)
Đặt \(\left(\dfrac{2}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow3x+2y+2z=7\)
\(C=\dfrac{4}{\dfrac{2}{a}+\dfrac{1}{b}}+\dfrac{9}{\dfrac{4}{a}+\dfrac{1}{c}}+\dfrac{4}{\dfrac{1}{b}+\dfrac{1}{c}}=\dfrac{4}{x+y}+\dfrac{9}{2x+z}+\dfrac{4}{y+z}\)
\(C\ge\dfrac{\left(2+3+2\right)^2}{x+y+2x+z+y+z}=\dfrac{49}{7}=7\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(\left(a;b;c\right)=\left(2;1;1\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2021
Đặt \(\left(a+1;b+1;c+1\right)=\left(x;y;z\right)\Rightarrow1\le x\le y\le z\le2\)
\(B=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}+3\) (1)
Do \(x\le y\le z\Rightarrow\left(z-y\right)\left(y-x\right)\ge0\)
\(\Leftrightarrow xy+yz\ge y^2+zx\)
\(\Leftrightarrow\dfrac{x}{z}+1\ge\dfrac{y}{z}+\dfrac{x}{y}\)
Tương tự: \(1+\dfrac{z}{x}\ge\dfrac{y}{x}+\dfrac{z}{y}\)
Cộng vế: \(2+\dfrac{x}{z}+\dfrac{z}{x}\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{y}{x}\) (2)
Từ (1); (2) \(\Rightarrow B\le2\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+5\)
Đặt \(\dfrac{z}{x}=t\Rightarrow1\le t\le2\)
\(\Rightarrow B\le2\left(t+\dfrac{1}{t}\right)+5=\dfrac{2t^2+2}{t}+5=\dfrac{2t^2+2}{t}-5+10\)
\(\Rightarrow B\le\dfrac{2t^2-5t+2}{t}+10=\dfrac{\left(t-2\right)\left(2t-1\right)}{t}+10\le10\)
\(B_{max}=10\) khi \(t=2\) hay \(\left(a;b;c\right)=\left(0;0;1\right);\left(0;1;1\right)\)
AT
27 tháng 7 2021
\(P=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\) (BĐT Cauchy Schwarz)
\(=\dfrac{9}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{7}{ab+bc+ca}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2ac+2bc}+\dfrac{7}{ab+bc+ca}\)
\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)
Ta có: \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\) .Thế vào biểu thức
\(\Rightarrow P\ge9+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)
\(\Rightarrow P_{min}=30\) khi \(a=b=c=\dfrac{1}{3}\)
14 tháng 3 2021
Áp dụng bđt Schwarz ta có:
\(P=\dfrac{a^4}{2ab+3ac}+\dfrac{b^4}{2cb+3ab}+\dfrac{c^4}{2ac+3bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(a^2+b^2+c^2\right)}=\dfrac{1}{5}\).
Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\dfrac{\sqrt{3}}{3}\).