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Áp dụng bất đẳng thức \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) với \(x=a^2+2bc;y=b^2+2ac;z=c^2+2ab\)
Ta có : \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ac\right)}=\frac{9}{\left(a+b+c\right)^2}\)
\(\Rightarrow\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9\)( Vì a + b + c = 1)
\(VT\ge\frac{9}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
Cauchy Schwars
\(M\ge\frac{\left(1+1+1\right)^2}{\left(a+b+c\right)^2}=\frac{9}{\left(a+b+c\right)^2}\ge9\Rightarrow M_{min}=9\Leftrightarrow a=b=c=\frac{1}{3}\)
\(M=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{\left(a+b+c\right)^2}\ge9\)
Dau '=' xay ra khi \(a=b=c=\frac{1}{3}\)
Vay \(M_{min}=9\)
Câu này lớp 7 tớ có làm. Cũng như cái mà gọi là áp dụng t/c dãy tỉ số bằng nhau và tỉ lệ thức. mình tính ra dc a, b. c rồi.
a,b,c khác nhau đôi một nghĩa là từng cặp số khác nhau ,là:
+a khác b
+b khác c
+c khác a
\(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0\)
Suy ra: \(ab==-\left(bc+ac\right)=-bc-ac\)
\(bc=-\left(ab+ac\right)=-ab-ac\)
\(ac=-\left(ab+bc\right)=-ab-bc\)
Nên \(a^2+2ab=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự,ta cũng có: \(b^2+2ac=\left(b-a\right)\left(b-c\right)\)
\(c^2+2ab=\left(c-a\right)\left(c-b\right)\)
Vậy \(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Rightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-ac-bc\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)
\(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự: \(b^2+2ac=\left(b-c\right)\left(b-a\right)\)
\(c^2+2ab=\left(a-c\right)\left(b-c\right)\)
\(B=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}+\frac{ca+1}{\left(b-a\right)\left(b-c\right)}+\frac{ab+1}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}-\frac{ca+1}{\left(a-b\right)\left(b-c\right)}+\frac{ab+1}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-c\right)+\left(ab+1\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-c\right)+\left(ab+1\right)\left(a-b\right)\)
\(=\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-b\right)-\left(ca+1\right)\left(b-c\right)+\left(ab+1\right)\left(a-b\right)\)
\(=\left(b-c\right)\left(bc+1-ca-1\right)+\left(a-b\right)\left(ab+1-ca-1\right)\)
\(=\left(b-c\right)\left(bc-ca\right)+\left(a-b\right)\left(ab-ca\right)\)
\(=\left(b-c\right)c\left(b-a\right)+\left(a-b\right)a\left(b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
Vậy B = 1