Cho a+b-c=0 đặt A=\(\dfrac{4bc-a^2}{-bc+2a^2}\)
B=\(\dfrac{4ac-b^2}{2b^2-ac}\) , C=\(\dfrac{4ab-c^2}{ab+2c^2}\)
CM:A.B.C=1
K
Khách
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Những câu hỏi liên quan
29 tháng 7 2017
a/ \(\frac{4bc-a^2}{bc+2a^2}.\frac{4ab-c^2}{ab+2c^2}.\frac{4ac-b^2}{ac+2b^2}\)
\(=\frac{4bc-\left(b+c\right)^2}{bc+2\left(b+c\right)^2}.\frac{4\left(-b-c\right)b-c^2}{\left(-b-c\right)b+2c^2}.\frac{4\left(-b-c\right)c-b^2}{\left(-b-c\right)c+2b^2}\)
\(=\frac{-\left(b-c\right)^2}{\left(c+2b\right)\left(b+2c\right)}.\frac{-\left(c+2b\right)^2}{-\left(b-c\right)\left(b+2c\right)}.\frac{-\left(b+2c\right)^2}{\left(b-c\right)\left(c+2b\right)}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 3 2021
\(VT\ge\dfrac{1}{\left(a^2+1\right)-1}+\dfrac{1}{\left(b^2+1\right)-1}+\dfrac{1}{\left(c^2+1\right)-1}+4-\dfrac{4}{ab+1}+4-\dfrac{4}{bc+1}+4-\dfrac{4}{ca+1}\)
\(VT\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{4}{ab+1}-\dfrac{4}{bc+1}-\dfrac{4}{ca+1}+12\)
Mặt khác \(a;b;c\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab+1\ge a+b\) (và tương tự...)
\(\Rightarrow VT\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+12\)
\(VT\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+1+1+1+9\)
\(VT\ge\left(\dfrac{2}{a+b}-1\right)^2+\left(\dfrac{2}{b+c}-1\right)^2+\left(\dfrac{2}{c+a}-1\right)^2+9\ge9\)
KN
6 tháng 3 2021
* Vì \(a,b\ge1\)nên \(\left(a-1\right)\left(b-1\right)\ge0\Leftrightarrow ab+1\ge a+b\)
Một cách tương tự: \(bc+1\ge b+c;ca+1\ge c+a\)
Với mọi số thực \(a\ge1\) ta luôn có: \(\left(a-1\right)^2\ge0\Leftrightarrow a^2\ge2a-1\Leftrightarrow\frac{1}{2a-1}\ge\frac{1}{a^2}\)
Tương tự: \(\frac{1}{2b-1}\ge\frac{1}{b^2};\frac{1}{2c-1}\ge\frac{1}{c^2}\)
Từ đó suy ra \(VT\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{4ab}{ab+1}+\frac{4bc}{bc+1}+\frac{4ca}{ca+1}\)\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+4-\frac{4}{ab+1}+4-\frac{4}{bc+1}+4-\frac{4}{ca+1}\)\(\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}-\frac{4}{ab+1}-\frac{4}{bc+1}-\frac{4}{ca+1}+12\)\(\ge\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}-\frac{4}{a+b}-\frac{4}{b+c}-\frac{4}{c+a}+12\)\(=\left(\frac{2}{a+b}-1\right)^2+\left(\frac{2}{b+c}-1\right)^2+\left(\frac{2}{c+a}-1\right)^2+9\ge9\)
Đẳng thức xảy ra khi a = b = c = 1
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2021
\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)
Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)
Cộng vế với vế:
\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 12 2022
\(A=\dfrac{x-4+5}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5}{\sqrt{x}-2}=\sqrt{x}+2+\dfrac{5}{\sqrt{x}-2}\)
\(=\sqrt{x}-2+\dfrac{5}{\sqrt{x}-2}+4\ge2\sqrt{\dfrac{5\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+4=4+2\sqrt{5}\)
\(A_{min}=4+2\sqrt{5}\) khi \(9+4\sqrt{5}\)
b.
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{l}{z}\right)\Rightarrow xyz=1\)
\(B=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)
\(B_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\Rightarrow a=b=c=1\)
4 tháng 12 2018
\(A=\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\)
\(\Leftrightarrow2A=\dfrac{2a^2}{2a^2+bc}+\dfrac{2b^2}{2b^2+ac}+\dfrac{2c^2}{2c^2+ab}\)
\(=1-\dfrac{bc}{2a^2+bc}+1-\dfrac{ac}{2b^2+ac}+1-\dfrac{ab}{2c^2+ab}\)
\(=3-\dfrac{bc}{2a^2+bc}-\dfrac{ac}{2b^2+ac}-\dfrac{ab}{2c^2+ab}\)
CM: \(P=\dfrac{bc}{2a^2+bc}+\dfrac{ac}{2b^2+ac}+\dfrac{ab}{2c^2+ab}\ge1\)
Thật vậy:
\(P\ge\dfrac{\left(ab+bc+ac\right)^2}{2a^2bc+b^2c^2+2b^2ac+a^2c^2+2c^2ab+a^2b^2}\)
\(=\dfrac{\left(ab+bc+ac\right)^2}{a^2bc+a^2bc+b^2c^2+b^2ac+b^2ac+a^2c^2+c^2ab+c^2ab+a^2b^2}\)
\(=\dfrac{\left(ab+bc+ac\right)^2}{ab\left(ac+bc+ab\right)+bc\left(ab+bc+ac\right)+ac\left(ab+bc+ac\right)}\)
\(=1\)
\(2A=3-P\le3-1=2\)
\(2A\le2\Leftrightarrow A\le1\)
\("="\Leftrightarrow a=b=c\)
25 tháng 5 2018
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=> bc+ac+ab=0
ta có
\(bc+ac=-ab\)
<=> \(\left(bc+ac\right)^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)
tương tự
\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)
\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)
thay vào E ta đc
\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)
=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)
N
12 tháng 7 2017
M=\(\left(x_1+x_2\right)^2-2x_1.x_2+\left(y_1+y_2\right)^2-2y_1.y_2\)
Áp dụng định lý viettel :( :v )
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\);\(\left\{{}\begin{matrix}y_1+y_2=-\dfrac{b}{c}\\y_1y_2=\dfrac{a}{c}\end{matrix}\right.\)
\(M=\dfrac{b^2}{a^2}-\dfrac{2c}{a}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}=\dfrac{b^2-4ac}{a^2}+\dfrac{b^2-4ac}{c^2}+2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)
\(\ge2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge4\)
Dấu = xảy ra: \(\left\{{}\begin{matrix}a=c\\b^2=4ac\end{matrix}\right.\)\(\Leftrightarrow b^2=4a^2=4c^2\)
12 tháng 7 2017
@_@ 
đưa thẳng câu hỏi luôn đi ; nói như zầy chưa nghỉ ra câu trả lời ; chống mặt chết trước rồi
