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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
+ \(2a+b+c=\left(a+b\right)+\left(a+c\right)\)
\(\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) ( theo AM-GM )
\(\Rightarrow\left(2a+b+c\right)^2\ge4\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow b=c\)
+ Tương tự : \(\frac{1}{\left(2b+c+a\right)^2}\le\frac{1}{4\left(a+b\right)\left(b+c\right)}\). Dấu "=" xảy ra <=> a = c
\(\frac{1}{\left(2c+a+b\right)^2}\le\frac{1}{4\left(a+c\right)\left(b+c\right)}\). Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Do đó : \(P\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(b+c\right)}+\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\)
\(\Rightarrow P\le\frac{1}{2}\cdot\frac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}\)\(=8abc\)
\(\Rightarrow P\le\frac{a+b+c}{16abc}\)
+ \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\). Dấu :=" xảy ra \(\Leftrightarrow a=b\)
\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\). Dấu "=" xảy ra <=> b = c
\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\). Dấu "=" xảy ra <=> c = a
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow3\ge\frac{a+b+c}{abc}\) \(\Rightarrow a+b+c\le3abc\)
\(\Rightarrow P\le\frac{3abc}{16abc}=\frac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\Rightarrow\left(a+b+c\right)\left(ab+ac+bc\right)-abc=0\Rightarrow\left(a+b\right)\left(ab+ac+bc\right)+abc+ac^2+bc^2-abc=0\Rightarrow\left(a+b\right)\left(ab+ac+bc\right)+c^2\left(a+b\right)=0\Rightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\Rightarrow\left[{}\begin{matrix}a+b=0\\a+c=0\\b+c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=-b\\c=-a\\b=-c\end{matrix}\right.\)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
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 $a=b=c$
Bài 1:
a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)
b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)
c: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)
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 \(a=b=c=\dfrac{1}{2}\)
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