Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng BĐT AM-GM ta có:
\(\dfrac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\dfrac{1}{2}\ge\sqrt[3]{abc}\Rightarrow\dfrac{1}{8}\ge abc\)
Áp dụng BĐT Holder ta có:
\(B=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(\ge\left(\sqrt[3]{3\cdot3\cdot3}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\right)^3\)
\(=\left(3+2\sqrt[3]{\dfrac{1}{abc}}\right)^3\ge\left(3+2\sqrt[3]{\dfrac{1}{\dfrac{1}{8}}}\right)^3=343\)
Khi \(a=b=c=\dfrac{1}{2}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{\left(1-a\right)^2}+\dfrac{1-a}{8}+\dfrac{1-a}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{\left(1-b\right)^2}+\dfrac{1-b}{8}+\dfrac{1-b}{8}\ge\dfrac{3b}{4}\\\dfrac{c^3}{\left(1-c\right)^2}+\dfrac{1-c}{8}+\dfrac{1-c}{8}\ge\dfrac{3c}{4}\end{matrix}\right.\)
\(\Rightarrow P+\dfrac{6-2\left(a+b+c\right)}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow P\ge\dfrac{1}{4}\)
Vậy \(P_{min}=\dfrac{1}{4}\)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
Áp dụng BĐT Cô-si cho các số dương ta có:
(2a+b+c)2 = \(\left[\left(a+b\right)+\left(a+c\right)\right]^2\) \(\ge\) 4(a+b)(a+c)
\(\Rightarrow\) \(\dfrac{1}{\left(2a+b+c\right)^2}\) \(\le\) \(\dfrac{1}{4\left(a+b\right)\left(a+c\right)}\)
Tương tự : \(\dfrac{1}{\left(2b+c+a\right)^2}\) \(\le\) \(\dfrac{1}{4\left(b+c\right)\left(b+a\right)}\)
\(\dfrac{1}{\left(2c+a+b\right)^2}\) \(\le\) \(\dfrac{1}{4\left(c+b\right)\left(c+a\right)}\)
Cộng theo vế 3 đẳng thức trên
\(\dfrac{1}{\left(2a+b+c\right)^2}\)+\(\dfrac{1}{\left(2b+c+a\right)^2}\)+\(\dfrac{1}{\left(2c+a+b\right)^2}\) \(\le\)\(\dfrac{1}{4}\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(b+c\right)\left(b+a\right)}+\dfrac{1}{\left(c+b\right)\left(c+a\right)}\right)\)
=\(\dfrac{1}{4}\left(\dfrac{b+c+a+b+c+a}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\right)\)
Áp dụng BĐT Cô-si ta có:
\(a+b\ge2\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\) \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
\(\Rightarrow\) P \(\le\) \(\dfrac{a+b+c}{16abc}\) = \(\dfrac{1}{16}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\) \(\le16\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\) = \(\dfrac{3}{16}\)
\(\Rightarrow\) Pmax = \(\dfrac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow\) a = b = c = 1
Vậy Pmax = \(\dfrac{3}{16}\) \(\Leftrightarrow\) a = b = c = 1
Lời giải:
Áp dụng BĐT AM-GM:
\(\frac{1}{1+\frac{1}{a^3}}+\frac{1}{1+\frac{1}{b^3}}+\frac{1}{1+\frac{1}{c^3}}\geq 3\sqrt[3]{\frac{1}{(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})}}\)
\(\frac{\frac{1}{a^3}}{1+\frac{1}{a^3}}+\frac{\frac{1}{b^3}}{1+\frac{1}{b^3}}+\frac{\frac{1}{c^3}}{1+\frac{1}{c^3}}\geq 3\sqrt[3]{\frac{\frac{1}{a^3b^3c^3}}{(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})}}\)
Cộng theo vế:
\(\Rightarrow 3\geq 3.\frac{1+\frac{1}{abc}}{\sqrt[3]{(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})}}\)
\(\Rightarrow P=(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})\geq (1+\frac{1}{abc})^3\)
Mà theo AM-GM: \(6=a+b+c\geq 3\sqrt[3]{abc}\Rightarrow abc\leq 8\)
\(\Rightarrow P\geq (1+\frac{1}{abc})^3\geq (1+\frac{1}{8})^3=\frac{729}{512}\)
Vậy \(P_{\min}=\frac{729}{512}\Leftrightarrow a=b=c=2\)
HÌnh như là \(a+b+c\le\dfrac{3}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\dfrac{1}{2}\ge\sqrt[3]{abc}\)
Áp dụng BĐT Holder ta có:
\(A=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(\ge\left(\sqrt[3]{3^3}+\dfrac{1}{\sqrt[3]{abc}}+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)\(\ge\left(3+\dfrac{1}{\dfrac{1}{2}}+\dfrac{1}{\dfrac{1}{2}}\right)^3=343\)
Xảy ra khi \(a=b=c=\dfrac{1}{2}\)
cứ cho là a+b+c <=3/2
đã >=0 đâu mà G với M