Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(P=2\Sigma a+\Sigma\dfrac{1}{a}=\Sigma a+\Sigma a+\Sigma\dfrac{1}{a}\ge3.\sqrt[3]{\left(\Sigma a\right)^2.\Sigma\dfrac{1}{a}}\)
\(Q=\left(\Sigma a\right)^2.\Sigma\dfrac{1}{a}=\left(3+2\Sigma ab\right).\Sigma\dfrac{1}{a}=3\Sigma\dfrac{1}{a}+4\Sigma a+2\Sigma\dfrac{ab}{c}\ge3\Sigma\dfrac{1}{a}+6\Sigma a=3\left(\Sigma\dfrac{1}{a}+2\Sigma a\right)=3P\)\(\Rightarrow\)\(P\ge3\sqrt[3]{3P}\) \(\Leftrightarrow P^3\ge81P\Leftrightarrow P^2\ge81\left(P>0\right)\Leftrightarrow P\ge9\)
" = " \(\Leftrightarrow a=b=c=1\)
Vì $\large a,b,c \in\mathbb{N^*}$ và $\large a^2+b^2+c^2=3\Rightarrow \left\{\begin{matrix} a<\sqrt{3} & \\ b<\sqrt{3} & \\ c<\sqrt{3} & \end{matrix}\right.$
Ta chứng minh bất đẳng thức phụ sau:
Với $0 <x<\sqrt{3}$ thì $2x+\frac{1}{x} \ge x^2.\frac{1}{2}+\frac{5}{2}(*)$
Thật vậy $(*)$ $\large \Leftrightarrow (x-2)(x-1)^2 \le0$
Do $\large x<\sqrt{3}\Leftrightarrow x<2\Leftrightarrow (x-2)(x-1)^2<0$ (Luôn đúng)
Do đó bất đẳng thức được chứng minh
Dấu $"="$ xảy ra khi $x=1$
Trở lại bài toán:
Áp dụng BĐT $(*)$ ta được:
$\large 2a+\frac{1}{a}+2b+\frac{1}{b}+2c+\frac{1}{c}\ge\frac{1}{2}(a^2+b^2+c^2)+\frac{15}{2}=9$
Do $a^2+b^2+c^2=3$
Vậy $GTNN=9$
Dấu $"="$ xảy ra khi: $a=b=c=1$
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
Đặt \(A=\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\)
Vì \(a,b,c>0\)nên áp dụng bất đẳng thức Cô-si cho 3 số dương, ta được:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge3\sqrt[3]{\frac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right).64}}\)\(=3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\left(1\right)\)
Chứng minh tương tự, ta được:
\(\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{1+c}{8}+\frac{1+a}{8}\ge\frac{3b}{4}\left(2\right)\)
\(\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{1+a}{8}+\frac{1+b}{8}\ge\frac{3a}{4}\left(3\right)\)
Từ (1), (2), (3), ta được:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\)\(+\frac{1+a}{8}+\frac{1+b}{8}+\frac{1+c}{8}+\frac{1+a}{8}+\frac{1+b}{8}+\frac{1+c}{8}\)\(\ge\frac{3a}{4}+\frac{3b}{4}+\frac{3c}{4}\)
\(\Leftrightarrow A+\frac{1+a}{4}+\frac{1+b}{4}+\frac{1+c}{4}\ge\frac{3a}{4}+\frac{3b}{4}+\frac{3c}{4}\)
\(\Leftrightarrow A+\frac{1+a+1+b+1+c}{4}\ge\frac{3a+3b+3c}{4}\)
\(\Leftrightarrow A+\frac{3+a+b+c}{4}\ge\frac{3\left(a+b+c\right)}{4}\)
\(\Leftrightarrow A\ge\frac{3\left(a+b+c\right)}{4}-\frac{3-a-b-c}{4}\)
\(\Leftrightarrow A\ge\frac{3\left(a+b+c\right)-\left(a+b+c\right)}{4}-\frac{3}{4}\)
\(\Leftrightarrow A\ge\frac{2\left(a+b+c\right)}{4}-\frac{3}{4}\left(4\right)\)
Mặt khác, vì \(a,b,c>0\)nên áp dụng bất đẳng thức Cô-si cho 3 số dương, ta được:
\(a+b+c\ge3\sqrt[3]{abc}\)
Mà \(abc\ge1\Leftrightarrow\sqrt[3]{abc}\ge1\Leftrightarrow3\sqrt[3]{abc}\ge3\)
Do đó:
\(a+b+c\ge3\)
\(\Leftrightarrow2\left(a+b+c\right)\ge6\)
\(\Leftrightarrow\frac{2\left(a+b+c\right)}{4}\ge\frac{6}{4}=\frac{3}{2}\)
\(\Leftrightarrow\frac{2\left(a+b+c\right)}{4}-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\left(5\right)\)
Từ (4) và (5), ta được:
\(A\ge\frac{3}{4}\)(điều phải chứng minh)
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\abc=1\end{cases}}\Leftrightarrow a=b=c=1\)
Vậy \(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\)với \(a,b,c>0\)và \(abc\ge1\)
Với x,y>0x,y>0 đã cho, áp dụng bất đẳng thức Cô si ta có
\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+b}{x}+\dfrac{1+c}{y}\ge\dfrac{3a}{\sqrt[3]{xy}}(1+b)(1+c)a3+x1+b+y1+c≥3xy3a
Kỳ vọng rằng bất đẳng thức cần chứng minh trở thành đẳng thức khi a=b=c=1a=b=c=1, ta chọn x>0x>0 sao cho \dfrac{a^3}{\left(1+b\right)\left(1+c\right)}=\dfrac{1+b}{x}=\dfrac{1+c}{y}(1+b)(1+c)a3=x1+b=y1+c xảy ra khi a=b=c=1a=b=c=1, tức là \dfrac{1}{4}=\dfrac{2}{x}=\dfrac{2}{y}\Leftrightarrow x=y=841=
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 \(P_{\min}=\frac{729}{512}\Leftrightarrow a=b=c=2\)
\(P\ge3\sqrt[3]{\dfrac{abc\left(a^2+1\right)^2\left(b^2+1\right)^2\left(c^2+1\right)^2}{a^2b^2c^2\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}}=3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)
\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(\dfrac{a+b+c}{3}\right)^3}}=9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(a+b+c\right)^3}}\ge9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{2\left(a+b+c\right)^2}}\)
Theo nguyên lý Dirichlet, trong 3 số \(a^2;b^2;c^2\) luôn có ít nhất 2 số cùng phía so với \(\dfrac{4}{9}\)
Không mất tính tổng quát, giả sử đó là \(a^2;b^2\)
\(\Rightarrow\left(a^2-\dfrac{4}{9}\right)\left(b^2-\dfrac{4}{9}\right)\ge0\)
\(\Leftrightarrow a^2b^2+\dfrac{16}{81}\ge\dfrac{4}{9}a^2+\dfrac{4}{9}b^2\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\dfrac{13}{9}a^2+\dfrac{13}{9}b^2+\dfrac{65}{81}\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\left(c^2+1\right)\)
\(=\dfrac{13}{9}\left(a^2+b^2+\dfrac{4}{9}+\dfrac{1}{9}\right)\left(\dfrac{4}{9}+\dfrac{4}{9}+c^2+\dfrac{1}{9}\right)\)
\(\ge\dfrac{13}{9}\left(\dfrac{2}{3}a+\dfrac{2}{3}b+\dfrac{2}{3}c+\dfrac{1}{9}\right)^2\)
\(\Rightarrow P\ge9\sqrt[3]{\dfrac{\dfrac{13}{9}\left(\dfrac{2}{3}\left(a+b+c\right)+\dfrac{1}{9}\right)^2}{2\left(a+b+c\right)^2}}=9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9\left(a+b+c\right)}\right)^2}\)
\(P\ge9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9.2}\right)^2}=\dfrac{13}{2}\)
\(P_{min}=\dfrac{13}{2}\) khi \(a=b=c=\dfrac{2}{3}\)
Từ giả thiết \(2\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le\dfrac{2}{3}\)
\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)
Đặt \(Q=\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}\)
\(=\dfrac{a^2b^2c^2+\left(a^2b^2+b^2c^2+c^2a^2\right)+\left(a^2+b^2+c^2\right)+1}{abc}\)
\(\ge\dfrac{a^2b^2c^2+3\sqrt[3]{\left(a^2b^2c^2\right)^2}+3\sqrt[3]{a^2b^2c^2}+1}{abc}=\dfrac{\left(\sqrt[3]{a^2b^2c^2}+1\right)^3}{abc}\)
\(=\left(\dfrac{\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{abc}}+\dfrac{1}{\sqrt[3]{abc}}\right)^3=\left(\sqrt[3]{abc}+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)
\(=\left(\sqrt[3]{abc}+\dfrac{4}{9\sqrt[3]{abc}}+\dfrac{5}{9\sqrt[3]{abc}}\right)^3\ge\left(2\sqrt[]{\dfrac{4\sqrt[3]{abc}}{9\sqrt[3]{abc}}}+\dfrac{5}{9.\dfrac{2}{3}}\right)^3=\dfrac{2197}{216}\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{2197}{216}}=\dfrac{13}{2}\)