Cho a> 10, b >100, c >1000. Tìm MIn
P=\(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
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NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 8 2020
1.
\(P=\frac{x}{2}+\frac{1}{2x}+\frac{5x}{2}\ge2\sqrt{\frac{x}{4x}}+\frac{5}{2}.1=\frac{7}{2}\)
Dấu "=" xảy ra khi \(x=1\)
2.
\(P=\frac{a}{100}+\frac{1}{a}+\frac{b}{10000}+\frac{1}{b}+\frac{c}{1000^2}+\frac{1}{c}+\frac{99}{100}a+\frac{9999}{10000}b+\frac{999999}{1000000}c\)
\(P\ge2\sqrt{\frac{a}{100a}}+2\sqrt{\frac{b}{10000b}}+2\sqrt{\frac{c}{1000000c}}+\frac{99}{100}.10+\frac{9999}{10000}.100+\frac{999999}{1000000}.1000=...\)
Bạn tự bấm máy tính
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=10\\b=100\\c=1000\end{matrix}\right.\)
3.
\(VT=\frac{a^2+b^2}{ab}+\frac{8ab}{\left(a+b\right)^2}\ge\frac{\left(a+b\right)^2}{2ab}+\frac{8ab}{\left(a+b\right)^2}\ge2\sqrt{\frac{8ab\left(a+b\right)^2}{2ab\left(a+b\right)^2}}=4\)
Dấu "=" xảy ra khi \(a=b\)
NL
1
15 tháng 8 2017
Theo AM-GM có :
\(9a+\frac{1}{a}\ge2\sqrt{9a.\frac{1}{a}}=6\)
\(9b+\frac{1}{b}\ge2\sqrt{9b.\frac{1}{b}}=6\)
\(9c+\frac{1}{c}\ge2\sqrt{9c.\frac{1}{c}}=6\)
Cộng vế theo vế :
\(9\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge18\)
\(8\left(a+b+c\right)+\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)\ge18\)
\(\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)\ge10\)
Vậy \(Min=10\)
\(\Leftrightarrow a=b=c=\frac{1}{3}\)
NG
Cho a,b,c > 0 thỏa mãn a+b+c=1. Tìm Min \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}+\frac{1}{9abc}\)
1
28 tháng 5 2018
\(A\ge\frac{9}{a+2+b+2+c+2}+\frac{1}{9abc}\)
\(\Rightarrow A\ge\frac{9}{7}+\frac{1}{9abc}\)
Theo BĐT AM-GM ta có: \(1=a+b+c\ge3\sqrt[3]{abc}\)
\(\Rightarrow abc\le\frac{1}{27}\)
\(\Rightarrow\frac{1}{9abc}\ge3\)
Do đó ta có:
\(A\ge\frac{9}{7}+3=\frac{30}{7}\)
T
0
KN
17 tháng 4 2020
\(A=\text{∑}_{cyc}\frac{a}{a^2+1}+\frac{1}{9abc}=\text{∑}_{cyc}\frac{1}{a+\frac{1}{a}}+\frac{1}{9abc}\)
\(\ge\frac{9}{\text{∑}_{cyc}\left(a+\frac{1}{a}\right)}+\frac{1}{9abc}=P\)
Ta có \(P=\frac{9}{\frac{1}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+\frac{1}{9abc}\)(Vì a + b + c = 1)
\(\ge\frac{9}{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{9}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+\frac{1}{9abc}\)
\(=\frac{81}{10}.\frac{abc}{ab+bc+ca}+\frac{1}{9abc}\)
\(\Rightarrow P\ge2\sqrt{\frac{3}{ab+bc+ca}}-\frac{21}{10}\ge2\sqrt{\frac{3}{\frac{\left(a+b+c\right)^2}{3}}}-\frac{21}{10}=\frac{39}{10}\)
\(\Rightarrow A\ge P\ge\frac{39}{10}\)
Dấu "=" khi và chỉ khi a = b = c = \(\frac{1}{3}\)
24 tháng 10 2020
Ta có:
\(\frac{1}{a+2}+\frac{3}{b+4}\le1-\frac{2}{c+3}\)
\(\Rightarrow1-\frac{1}{a+2}\ge\frac{3}{b+4}+\frac{2}{c+3}\ge2\sqrt{\frac{6}{\left(b+4\right)\left(c+3\right)}}\)
\(\Leftrightarrow\frac{a+1}{a+2}\ge2\sqrt{\frac{6}{\left(b+4\right)\left(c+3\right)}}\left(1\right)\)
Tương tự : \(1-\frac{3}{b+4}\ge\frac{1}{a+2}+\frac{2}{c+3}\ge2\sqrt{\frac{2}{\left(a+2\right)\left(c+3\right)}}\Leftrightarrow\frac{b+1}{b+4}\ge2\sqrt{\frac{2}{\left(a+2\right)\left(c+3\right)}}\left(2\right)\)
và \(\frac{c+1}{c+3}\ge2\sqrt{\frac{3}{\left(a+2\right)\left(b+4\right)}}\left(3\right)\)
Từ 1,2,3 ta có:
\(\frac{a+1}{a+2}.\frac{b+1}{b+4}.\frac{c+1}{c+3}\ge\frac{48}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\Leftrightarrow Q\ge48\)
Vậy Min Q =48 khi a=1,b=5,c=3