Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì a và 1 là 2 số dương \(\Rightarrow a+1\ge2\sqrt{a}\) (bđt AM - GM)
Vì b và 1 là 2 số dương \(\Rightarrow b+1\ge2\sqrt{b}\)(bđt AM - GM)
Vì c và 1 là 2 số dương \(\Rightarrow c+1\ge2\sqrt{c}\)(bđt AM - GM)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge2\sqrt{a}.2\sqrt{b}.2\sqrt{c}=8\sqrt{abc}=8\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có ít nhất 2 số cùng phía so với 1
Không mất tính tổng quát, giả sử đó là a và b
\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)
\(\Leftrightarrow ab+1\ge a+b\)
\(\Leftrightarrow2\left(ab+1\right)\ge\left(a+1\right)\left(b+1\right)\)
\(\Rightarrow\dfrac{2}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{2}{2\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(\dfrac{1}{c}+1\right)\left(c+1\right)}=\dfrac{c}{\left(c+1\right)^2}\)
Lại có:
\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(\dfrac{a}{b}+1\right)}+\dfrac{1}{\left(ab+1\right)\left(\dfrac{b}{a}+1\right)}=\dfrac{1}{ab+1}\)
\(\Rightarrow P\ge\dfrac{1}{ab+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}=\dfrac{1}{\dfrac{1}{c}+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}\)
\(\Rightarrow P\ge\dfrac{c}{c+1}+\dfrac{c+1}{\left(c+1\right)^2}=\dfrac{c\left(c+1\right)+c+1}{\left(c+1\right)^2}=\dfrac{\left(c+1\right)^2}{\left(c+1\right)^2}=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
AM-GM 1 dòng thôi bạn :))
\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge2\sqrt{a}.2\sqrt{b}.2\sqrt{c}=8\sqrt{abc}=8\)
Dấu "=" khi a=b=c=1
Áp dụng BĐT AM - GM cho các số không âm , ta có :
\(\left\{{}\begin{matrix}a+1\ge2\sqrt{a}\\b+1\ge2\sqrt{b}\\c+1\ge2\sqrt{c}\end{matrix}\right.\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge2.2.2.\sqrt{abc}\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\left(a-1\right)^2\ge0\)
\(\Leftrightarrow a^2-2a+1\ge0\)
\(\Leftrightarrow\left(a^2+2a+1\right)-4a\ge0\)
\(\Leftrightarrow\left(a+1\right)^2\ge4a\)
TT\(\Rightarrow\left(b+1\right)^2\ge4b\)
\(\left(c+1\right)^2\ge4b\)
Nhân vế theo vế ta được \(\left[\left(a+1\right)\left(b+1\right)\left(c+1\right)\right]^2\ge64abc=64\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức Cauchy cho từng cặp số không âm (với \(a,b,c>0\)), ta có:
\(a^2+1\ge2a\) \(\left(1\right)\)
\(b^2+1\ge2b\) \(\left(2\right)\)
\(c^2+1\ge2c\) \(\left(3\right)\)
Nhân từng vế \(\left(1\right);\) \(\left(2\right)\) và \(\left(3\right)\), ta được:
\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge2a.2b.2c=8abc=8\) (do \(abc=1\))
Xảy ra đẳng thức trên khi và chỉ khi \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a+1\right)\left(b+1\right)\left(c+1\right)=abc+ac+bc+ab+a+b+c+1\)
Áp dụng BĐT thức Cô si cho 3 số , ta có:
\(a+b+c\ge3\sqrt[3]{abc}=3\)
\(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\)
\(\Rightarrow ab+bc+ca+a+b+c+2\ge3+3+2=8\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(a+b+1\right)\left(a^2+b^2\right)+\frac{4}{a+b}\)
\(\Rightarrow A\ge\left(a+b+1\right).2ab+\frac{4}{a+b}=2\left(a+b+1\right)+\frac{4}{a+b}\)
\(\Rightarrow A\ge\left(a+b\right)+\left(a+b\right)+\frac{4}{a+b}+2\)
\(\Rightarrow A\ge2\sqrt{ab}+2\sqrt{\left(a+b\right).\frac{4}{a+b}}+2\)
\(\Rightarrow A\ge2+4+2=8\)
"=" khi \(a=b=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ca\right)^2}{b\left(c+a\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)
\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).
ĐTXR \(\Leftrightarrow a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
thử bài bất :D
Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)
Hoàn toàn tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)
Cộng (*),(**),(***) vế theo vế ta được:
\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)
Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )
Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D