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\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)
\(\Leftrightarrow a\left(ab+ac+bc\right)+\left(b+c\right)\left(ab+ac+bc\right)-abc=0\)
\(\Leftrightarrow a\left(ab+ac+bc-bc\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow a^2\left(b+c\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow\left(a^2+ab+ac+bc\right)\left(b+c\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)
- Nếu \(a=-c\Rightarrow a^{2006}=c^{2006}\Rightarrow c^{2006}-a^{2006}=0\Rightarrow P=0\)
- Nếu \(a=-b\Rightarrow a^{2004}=b^{2004}\Rightarrow a^{2004}-b^{2004}=0\Rightarrow P=0\)
- Nếu \(b=-c\Rightarrow b^{2005}=-c^{2005}\Rightarrow b^{2005}+c^{2005}=0\Rightarrow P=0\)
Vậy \(P=0\)
Từ giả thiết suy ra: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a+b+c}\right)=0\)
\(\Rightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\Rightarrow\) (a + b)[c(a + b + c) + ab] = 0
\(\Rightarrow\) (a + b)(ac + ab + bc + c2) = 0
\(\Rightarrow\) (a + b)(b + c)(a + c) = 0
P = (a2004 - b2004)(b2005 + c2005)(c2006 - a2006)
= (a + b)(b + c)(a + c) = 0
Bạn tham khảo :
Ta có :
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3=1\)
\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2=0\)
\(\Rightarrow abc\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2\right)=abc.0\)
\(\Rightarrow a^2b+b^2c+a^2c+b^2a+c^2a+c^2b+2abc=0\)
\(\Rightarrow\left(a^2b+ab^2\right)+\left(b^2c+abc\right)+\left(a^2c+abc\right)+\left(c^2a+c^2b\right)=0\)
\(\Rightarrow ab\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Rightarrow\left(ab+bc+ac+c^2\right)\left(a+b\right)=0\)
\(\Rightarrow\left[\left(ab+bc\right)+\left(ac+c^2\right)\right]\left(a+b\right)=0\)
\(\Rightarrow\left[b\left(a+c\right)+c\left(a+c\right)\right]\left(a+b\right)=0\)
\(\Rightarrow\left(a+c\right)\left(b+c\right)\left(a+b\right)=0\)
TH1 : \(a+c=0\)
\(\Rightarrow a=-c\)
\(\Rightarrow c^{2006}=a^{2006}\)
\(\Rightarrow P=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)\left(c^{2006}-a^{2006}\right)\)
\(=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)0\)
\(=0\)
CMTT đều có \(P=0\)
Vậy ...
a) vì ab > 0 nên chia cả hai vế Bất đẳng thức cho \(\sqrt{ab}\) ta được
\(\sqrt{\dfrac{c\left(a-c\right)}{ab}}+\sqrt{\dfrac{c\left(b-c\right)}{ab}}\le1\)
Áp dụng Bất đẳng thức Cauchy cho hai số
\(\Rightarrow\sqrt{\dfrac{c}{b}\left(\dfrac{a-c}{a}\right)}+\sqrt{\dfrac{c}{a}\left(\dfrac{b-c}{b}\right)}\le\dfrac{1}{2}\left(\dfrac{c}{b}+\dfrac{a-c}{a}\right)+\dfrac{1}{2}\left(\dfrac{c}{a}+\dfrac{b-c}{b}\right)=1\)
vậy nên ta có đpcm
\(\frac{2005}{\sqrt{2006} }+\frac{2006}{\sqrt{2005} }>\sqrt{2005}+\sqrt{2006} \)
<=>\(2005\sqrt{2005}+2006\sqrt{2006}>2005\sqrt{2006}+2006\sqrt{2005} \)
<=>\(\sqrt{2006}<\sqrt{2005} \)
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
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a+b+c}\right)=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\times\dfrac{ac+bc+c^2+ab}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
\(\Rightarrow N=0\)
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
a) phương trình \(x^3-3x^2+1\) có 3 nghiệm thực phân biệt là a,b,c(đề bài). Áp dụng Định lí Vi-ét cho đa thức bậc 3 ta có:\(\left\{{}\begin{matrix}a+b+c=3\\ab+bc+ac=0\\a.b.c=-1\end{matrix}\right.\)
ta có
a+b+c=3
<=>\(\left(a+b+c\right)^2=9\)
<=>\(a^2+b^2+c^2+2ab+2bc+2ac=9\)
<=>\(a^2+b^2+c^2=9\)
<=>\(\left(a^2+b^2+c^2\right)^2=81\)
<=>\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=81\)(1)
ta có ab+bc+ac=0
<=>\(\left(ab+bc+ac\right)^2=0\)
<=>\(a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=0\)
<=>\(a^2b^2+b^2c^2+a^2c^2-2.1.3=0\)
<=>\(a^2b^2+b^2c^2+a^2c^2=6\)(2)
Thay (2) vào (1) ta có \(a^4+b^4+c^4+2.6=81\)
<=>\(a^4+b^4+c^4=69\)
b) \(\dfrac{a+1}{\left(b+c\right)\left(1-a\right)+1}=\dfrac{a+1}{\left(3-a\right)\left(1-a\right)+1}=\dfrac{a+1}{3+a^2-4a+1}=\dfrac{a+1}{a^2-4a+4}=\dfrac{a+1}{\left(a-2\right)^2}\)
cmtt =>\(B=\dfrac{a+1}{\left(a-2\right)^2}+\dfrac{b+1}{\left(b-2\right)^2}+\dfrac{c+1}{\left(c-2\right)^2}\)=\(\dfrac{1}{a-2}+\dfrac{1}{b-2}+\dfrac{1}{c-2}+3\left[\dfrac{1}{\left(a-2\right)^2}+\dfrac{1}{\left(b-2\right)^2}+\dfrac{1}{\left(c-2\right)^2}\right]\)=\(\dfrac{3\left[\left(a-2\right)\left(b-2\right)\right]^2+3\left[\left(b-2\right)\left(c-a\right)\right]^2+3\left[\left(c-2\right)\left(a-2\right)\right]^2}{\left[\left(a-2\right)\left(b-2\right)\left(c-2\right)\right]^2}\)
đặt t=(a-2)(b-2);u=(b-2)(c-2);v=(c-2)(a-2) =>t+u+v=0
B thành \(\dfrac{3\left(t^2+u^2+v^2\right)}{t.u.v}\) bạn biến đổi để xuất hiện t+u+v
=>B=\(\dfrac{3\left(t+u+v\right)^2-6\left(t.u+u.v+t.v\right)}{t.u.v}=\dfrac{-6.\left(a-2\right)\left(b-2\right)\left(c-2\right)\left(a-2+b-2+c-2\right)}{t.u.v}=\dfrac{18}{\left(a-2\right)\left(b-2\right)\left(c-2\right)}\)
(a-2)(b-2)(c-2)= abc-2(ab+bc+ac)+4(a+b+c)-8=12-9=3
Vậy B=3