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\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a+b+c}\right)=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\times\dfrac{ac+bc+c^2+ab}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
\(\Rightarrow N=0\)
bài 2: ta có : \(Q=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-\left(1-a\right)}\right)\left(\sqrt{\dfrac{1}{a^2}-1}-\dfrac{1}{a}\right).\sqrt{a^2-2a+1}\)
\(\Leftrightarrow Q=\left(\dfrac{\sqrt{1+a}\sqrt{1-a}+1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\dfrac{\sqrt{1-a^2}}{a}-\dfrac{1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{1-a^2}-1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{\sqrt{1-a^2}+1}{a}\right)\left(\dfrac{\sqrt{1-a^2}-1}{a}\right)\left(1-a\right)\) \(\Leftrightarrow Q=\left(\dfrac{1-a^2-1}{a^2}\right)\left(1-a\right)=a-1\)b) ta có : \(Q^3-Q=\left(a-1\right)\left(\left(a-1\right)^2-1\right)=a\left(a-1\right)\left(a-2\right)\)
mà ta có : \(\left\{{}\begin{matrix}a>0\\a-1< 0\\a-2< 0\end{matrix}\right.\Rightarrow a\left(a-1\right)\left(a-2\right)>0\) \(\Rightarrow Q^3-Q>0\Leftrightarrow Q^3>Q\)
vậy \(Q^3>Q\)
Nguyễn Huy TúAkai HarumaLightning FarronNguyễn Thanh Hằngsoyeon_Tiểubàng giảiMashiro ShiinaVõ Đông Anh Tuấn
Hoàng Lê Bảo NgọcTrần Việt Linh
cứu tôi với
ĐKXĐ: \(abc\ne0\)
\(a^3+b^3+3ab\left(a+b\right)+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: \(a+b+c=0\)
\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)
TH2: \(a=b=c\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Bài 1:
\(P=(x+1)\left(1+\frac{1}{y}\right)+(y+1)\left(1+\frac{1}{x}\right)\)
\(=2+x+y+\frac{x}{y}+\frac{y}{x}+\frac{1}{x}+\frac{1}{y}\)
Áp dụng BĐT Cô-si:
\(\frac{x}{y}+\frac{y}{x}\geq 2\)
\(x+\frac{1}{2x}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
\(y+\frac{1}{2y}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
Áp dụng BĐT SVac-xơ kết hợp với Cô-si:
\(\frac{1}{2x}+\frac{1}{2y}\geq \frac{4}{2x+2y}=\frac{2}{x+y}\geq \frac{2}{\sqrt{2(x^2+y^2)}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Cộng các BĐT trên :
\(\Rightarrow P\geq 2+2+\sqrt{2}+\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
Vậy \(P_{\min}=4+3\sqrt{2}\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)
Bài 2:
Áp dụng BĐT Svac-xơ:
\(\frac{1}{a+3b}+\frac{1}{b+a+2c}\geq \frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)
\(\frac{1}{b+3c}+\frac{1}{b+c+2a}\geq \frac{4}{2b+4c+2a}=\frac{2}{b+2c+a}\)
\(\frac{1}{c+3a}+\frac{1}{c+a+2b}\geq \frac{4}{2c+4a+2b}=\frac{2}{c+2a+b}\)
Cộng theo vế và rút gọn :
\(\Rightarrow \frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\geq \frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\Rightarrow\left(a+b+c\right)\left(ab+ac+bc\right)-abc=0\Rightarrow\left(a+b\right)\left(ab+ac+bc\right)+abc+ac^2+bc^2-abc=0\Rightarrow\left(a+b\right)\left(ab+ac+bc\right)+c^2\left(a+b\right)=0\Rightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\Rightarrow\left[{}\begin{matrix}a+b=0\\a+c=0\\b+c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=-b\\c=-a\\b=-c\end{matrix}\right.\)TH1: nếu a=-b
P=(a2017+b2017)(b2018-c2018)=(-b2017+b2017)(b2018-c2018)=0
TH2: nếu b=-c
P=(a2017+b2017)(b2018-c2018)=(a2017+b2017)((-c)2018-c2018)=0
Còn một TH nữa thì bạn ghi thiếu đề rồi
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 \(P_{\min}=\frac{729}{512}\Leftrightarrow a=b=c=2\)
Lời giải:
Ta có:
\(\text{VT}=\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\) (1)
Thay \(1=a+b+c\) kết hợp với bất đẳng thức AM-GM:
\((a+1)(b+1)(c+1)=(a+a+b+c)(b+a+b+c)(c+a+b+c)\)
\(=[(a+b)+(a+c)][(b+c)(b+a)][(c+a)+(c+b)]\)
\(\geq 2\sqrt{(a+b)(a+c)}.2\sqrt{(b+c)(b+a)}.2\sqrt{(c+a)(c+b)}\)
\(\Leftrightarrow (a+1)(b+1)(c+1)\geq 8(a+b)(b+c)(c+a)\)
Tiếp tục áp dụng AM-GM:
\((a+b)(b+c)(c+a)\geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}=8abc\)
Suy ra \((a+1)(b+1)(c+1)\geq 64abc\) (2)
Từ (1);(2) ta có \(\text{VT}\geq 64\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\)
Từ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)
\(\Leftrightarrow a^2b+abc+a^2c+b^2a+b^2c+abc+bc^2+ac^2=0\)
\(\Leftrightarrow ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(ab+ac+bc+c^2\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left[a\left(b+c\right)+c\left(b+c\right)\right]\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+c\right)\left(b+c\right)\left(a+b\right)=0\)
Thay vào từng TH suy ra M=0