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Đầu tiên ta cm:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)(tự cm)
Áp dụng:\(\Rightarrow\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}\)
Lại có:\(a^2+b^2+c^2+2ab+2bc+2ca=\left(a+b+c\right)^2\le1\)
\(\Rightarrow\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}\ge\dfrac{9}{1}=9\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)
\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)
\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
cho mình hỏi bạn biết làm chưa nếu rồi thì giúp mình được không ạ mình ko biết làm
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow ab+bc+ca=0\)
\(C=\dfrac{a^2}{a^2+2bc}+\dfrac{b^2}{b^2+2ac}+\dfrac{c^2}{c^2+2ab}\)
\(=\dfrac{a^2}{a^2+bc-ac-ab}+\dfrac{b^2}{b^2+ac-ba-bc}+\dfrac{c^2}{c^2+ab-ca-cb}\)
\(=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=-\left(\dfrac{a^2}{\left(a-b\right)\left(c-a\right)}+\dfrac{b^2}{\left(a-b\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\right)\)
\(=-\left(\dfrac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)\)
\(=-\left(\dfrac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)=1\)
Ta có :
\(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}=\frac{3^2}{\left(a+b+c\right)^2}\)
\(A\ge\frac{9}{3^2}=1\)Dấu "=" khi a=b=c= 1
Lời giải:
Xét tử :
\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}=\frac{a^2}{a^2+bc+(-ab-ac)}+\frac{b^2}{b^2+ac+(-ab-bc)}+\frac{c^2}{c^2+ab+(-bc-ac)}\)
\(=\frac{a^2}{a(a-b)-c(a-b)}+\frac{b^2}{b(b-c)-a(b-c)}+\frac{c^2}{c(c-a)-b(c-a)}\)
\(=\frac{a^2}{(a-c)(a-b)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)}\)
\(=\frac{a^2(c-b)+b^2(a-c)+c^2(b-a)}{(a-b)(b-c)(c-a)}\)
\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)
Xét mẫu (tương tự bên tử)
\(\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}=\frac{bc}{(a-c)(a-b)}+\frac{ac}{(b-a)(b-c)}+\frac{ab}{(c-a)(c-b)}\)
\(=\frac{bc(c-b)+ac(a-c)+ab(b-a)}{(a-b)(b-c)(c-a)}=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(a-b)(b-c)(c-a)}\)
\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)
Do đó:
\(A=\frac{1}{1}=1\)
Ta có : 1/M=a2+2bc+b2+2ac+c2+2ab
=(a+b+c)2 ➝ M=1/(a+b+c)2
mik nghĩ là thế
Có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow ab+bc+ac=0\)
\(1\Leftrightarrow a^2+2bc=a^2+bc-ab-ac\)
\(\Leftrightarrow a^2+2bc=a\left(a-b\right)-c\left(a-b\right)\)
\(\Leftrightarrow a^2+2bc=\left(a-b\right)\left(b-c\right)\)
\(2\Leftrightarrow b^2+2ac=b^2+ac-ab-bc\)
\(\Leftrightarrow b^2+2ac=b\left(b-c\right)-a\left(b-c\right)\)
\(\Leftrightarrow b^2+2ac=\left(b-c\right)\left(b-a\right)\)
\(3.c^2+2ab=c^2+ab-bc-ac\)
\(\Leftrightarrow c^2+2ab=c\left(c-b\right)-a\left(c-b\right)\)
\(\Leftrightarrow c^2+2ab=\left(c-a\right)\left(c-b\right)\)
\(\Rightarrow M=\dfrac{1}{\left(a-b\right)\left(a-c\right)}+\dfrac{1}{\left(b-a\right)\left(b-c\right)}+\dfrac{1}{\left(c-a\right)\left(c-b\right)}\)
\(\Rightarrow M=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)
\(\Rightarrow M=\dfrac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\Rightarrow M=0\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{a^2+2bc+b^2+2ac+c^2+2ab}\)
\(=\dfrac{3^2}{\left(a+b+c\right)^2}=\dfrac{9}{\left(a+b+c\right)^2}=9\left(a+b+c\le1\right)\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)