Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
bai nay t lam roi vao trang chu cua nick thangbnsh cua t keo xuong tim la thay
Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
Từ gt , ta có :
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\left(a+b\right)c\left(a+b+c\right)=-\left(a+b\right)ab\)
\(\Rightarrow0=\left(a+b\right)\left(ca+cb+c^2\right)-\left[-\left(a+b\right)ab\right]=\left(a+b\right)\left(ca+cb+c^2+ab\right)=\left(a+b\right)\left(c+a\right)\left(c+b\right)\)
\(\Rightarrow a+b=0\) hoặc \(c+a=0\) . Gỉa sử \(a=-b\) thì \(a^{15}=-b^{15}\) nên \(a^{15}+b^{15}=0\)
\(\Rightarrow N=0\)
Áp dụng BĐT AM-GM: \(VT\le\sum\dfrac{1}{\sqrt{a^2+1}.\sqrt{2a}.2\sqrt{bc}}=\sum\dfrac{1}{2\sqrt{2}\sqrt{a^2+1}}\)
Ta đi chứng minh \(\dfrac{1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{b^2+1}}+\dfrac{1}{\sqrt{c^2+1}}\le\dfrac{3}{\sqrt{2}}\)
Giả sử c=max{a, b, c}.Suy ra \(c\ge1\) nên \(ab\le1\). Ta có bổ đề:
\(\dfrac{1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{b^2+1}}\le\dfrac{2}{\sqrt{1+ab}}\)(*)
#cm: Áp dụng Bunyakovsky: \(VT_{(*)} \)\(\le\sqrt{2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\right)}\)
Xét \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}-\dfrac{2}{ab+1}=\dfrac{\left(a-b\right)^2\left(ab-1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\le0\)
Nên \(VT_{(*)}\)\(\le\sqrt{2.\dfrac{2}{ab+1}}=\dfrac{2}{\sqrt{ab+1}}\), suy ra đpcm.
Do đó \(VT\le\dfrac{2}{\sqrt{ab+1}}+\dfrac{1}{\sqrt{c^2+1}}=2\sqrt{\dfrac{c}{c+1}}+\dfrac{1}{\sqrt{c^2+1}}\)
# cm: \(2\sqrt{\dfrac{c}{c+1}}+\dfrac{1}{\sqrt{c^2+1}}\le\dfrac{3}{\sqrt{2}}\)
\(\Leftrightarrow2\sqrt{2c\left(c^2+1\right)}+\sqrt{2c+2}\le3\sqrt{\left(c+1\right)\left(c^2+1\right)}\)
\(\Leftrightarrow8c^3+10c+2+8\sqrt{c\left(c+1\right)\left(c^2+1\right)}\le9\left(c^3+c^2+c+1\right)\)
hay \(8\sqrt{\left(c^2+c\right)\left(c^2+1\right)}\le c^3+9c^2-c+7\) ($)
Áp dụng BĐT AM-GM cho VT của ($):
\(8\sqrt{\left(c^2+c\right)\left(c^2+1\right)}\le4\left(2c^2+c+1\right)\) .Ta chứng minh
\(8c^2+4c+4\le c^3+9c^2-c+7\) hay \(\left(c-1\right)^2\left(c+3\right)\ge0\) (đúng)
Vậy ta có đpcm. Dấu = xảy ra khi a=b=c=1
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: \(a=b=c\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
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 \(P_{\min}=\frac{729}{512}\Leftrightarrow a=b=c=2\)
b,\(B=\sqrt{1+2014^2+\dfrac{2014^2}{2015^2}}+\dfrac{2014}{2015}\)
Ta có :\(\left(2014+1\right)^2=2014^2+1+2.2014\)
\(\Rightarrow2014^2+1=2015^2-2.2014\)
\(\Rightarrow B=\sqrt{2015^2-2.2014+\left(\dfrac{2014}{2015}\right)^2}+\dfrac{2014}{2015}\)
\(=\sqrt{\left(2015-\dfrac{2014}{2015}\right)^2}+\dfrac{2014}{2015}\)
\(=2015-\dfrac{2014}{2015}+\dfrac{2014}{2015}\)
\(=2015\)
Vậy B=2015
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a+b+c}\right)=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\times\dfrac{ac+bc+c^2+ab}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
\(\Rightarrow N=0\)