Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a^3+b^3+3a^2b+3b^2a\right)+c^3-3a^2b-3b^2a-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[a^2+b^2+2ab-ac-bc+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\left(1\right)\)
C/m : \(a^2+b^2+c^2-ab-bc-ac\ge0\)
Giả sử điều phải c/m là đúng , ta có :
\(a^2+b^2+c^2-ab-bc-ac\ge0\)
\(\Rightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)\ge0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) ( điều này luôn đúng )
\(\Rightarrow\) điều giả sử là đúng
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac\ge0\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow a+b+c=0\)
\(\Rightarrow a+b=-c;b+c=-a;a+c=-b\)
Lại có : \(A=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{a}\right)\)
\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)
\(=\dfrac{-abc}{abc}=-1\)
Vậy \(A=-1\)
Câu 1:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\2a^2+2b^2+2c^2-2ab-2ac-2bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\end{matrix}\right.\)
TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
\(M=\dfrac{\left(a+b\right)}{b}.\dfrac{\left(b+c\right)}{c}.\dfrac{\left(a+c\right)}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=\dfrac{-abc}{abc}=-1\)
TH2: \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\Leftrightarrow a=b=c\)
\(M=\left(\dfrac{a}{a}+1\right)\left(\dfrac{a}{a}+1\right)\left(\dfrac{a}{a}+1\right)=2.2.2=8\)
Lời giải:
Áp dụng BĐT AM-GM cho các số dương ta có:
\(\frac{a^3}{(a+1)(b+1)}+\frac{a+1}{8}+\frac{b+1}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
\(\frac{b^3}{(b+1)(c+1)}+\frac{b+1}{8}+\frac{c+1}{8}\geq 3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)
\(\frac{c^3}{(c+1)(a+1)}+\frac{c+1}{8}+\frac{a+1}{8}\geq 3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)
Cộng theo vế:
\(\Rightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{a+b+c+3}{4}\geq \frac{3}{4}(a+b+c)\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{3}{2}\geq \frac{9}{4}\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}\geq \frac{3}{4}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
A=\(\left(a+b\right)\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)
= \(\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{a}{b}+\dfrac{b}{a}\)
= \(2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)
Áp dụng BĐT cô si cho 2 số ta có
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}\)
⇔\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
⇔\(2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge4\)
⇔ A ≥4
=> Min A =4
dấu "=" xảy ra khi
\(\dfrac{a}{b}=\dfrac{b}{a}\)
⇔a2=b2
⇔a=b
vậy Min A =4 khi a=b
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(B=\frac{1}{(a+2b)(a+2c)}+\frac{1}{(b+2a)(b+2c)}+\frac{1}{(c+2a)(c+2b)}\)
\(\geq \frac{9}{(a+2b)(a+2c)+(b+2a)(b+2c)+(c+2a)(c+2b)}\)
\(\Leftrightarrow B\geq \frac{9}{(a^2+2ac+2ab+4bc)+(b^2+2bc+2ab+4ac)+(c^2+2bc+2ac+4ab)}\)
\(\Leftrightarrow B\geq \frac{9}{a^2+b^2+c^2+8(ab+bc+ac)}=\frac{9}{(a+b+c)^2+6(ab+bc+ac)}(*)\)
Theo hệ quả quen thuộc của BĐT Cô-si:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow (a+b+c)^2\geq 3(ab+bc+ac)\)
\(\Rightarrow 2(a+b+c)^2\geq 6(ab+bc+ac)(**)\)
Từ \((*); (**)\Rightarrow B\geq \frac{9}{(a+b+c)^2+2(a+b+c)^2}=\frac{3}{(a+b+c)^2}\geq \frac{3}{3^2}=\frac{1}{3}\)
(do \(a+b+c\leq 3)\)
Do đó: \(B_{\min}=\frac{1}{3}\)
Dấu bằng xảy ra khi \(a=b=c=1\)
-Hoặc có thể đề bạn sai
hiểu đơn giản thì có thể hiểu như sau:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc>abc\)
-Không có a;b;c thỏa mãn
Từ \(a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a+b+c=0\) hoặc \(a=b;b=c;c=a\) (bn tự chứng minh)
Với \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;a+c=-b\)
Ta có: \(A=\left(\dfrac{a}{b}+1\right).\left(\dfrac{b}{c}+1\right)\left(\dfrac{c}{a}+1\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
Với \(a=b;b=c;c=a\)
\(\Rightarrow A=\dfrac{a+b}{b}.\dfrac{b+c}{c}+\dfrac{c+a}{a}=\dfrac{2b}{b}.\dfrac{2c}{c}.\dfrac{2a}{a}=8\)
Áp dụng bđt AM-GM cho 2 số dương:
\(a^3+b^3+c^3\ge3abc\)
Dấu "=" xảy ra khi:
\(a=b=c\)
Khi đó:
\(\left\{{}\begin{matrix}\dfrac{a}{b}=1\\\dfrac{b}{c}=1\\\dfrac{a}{c}=1\end{matrix}\right.\) \(\Leftrightarrow\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{a}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a+b+c=0\) hoặc \(a=b=c\) (bn tự chứng minh)
+) \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;a+c=-b\)\(\Rightarrow A=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)
\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
+) \(a=b=c\Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)