Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt A= \(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\)
Đặt x = b + c - a, y = a + c - b, z =a + b -c
=>\(\left\{{}\begin{matrix}x+y=2c\\y+z=2a\\x+z=2b\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\end{matrix}\right.\)
\(\Leftrightarrow A=2\left(\dfrac{\dfrac{y+z}{2}}{x}+\dfrac{\dfrac{x+z}{2}}{y}+\dfrac{\dfrac{x+y}{2}}{z}\right)\)
\(\Leftrightarrow A=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)
\(\Leftrightarrow A=\dfrac{y}{x}+\dfrac{x}{y}+\dfrac{z}{x}+\dfrac{x}{z}+\dfrac{z}{y}+\dfrac{y}{z}\)
Theo bất đẳng thức Cô -si luôn đúng với m, n \(\ge0\)
=> \(m+n\ge2\sqrt{m.n}\) . Dấu '=' xảy ra kh m = n
=> Ta có : \(\left\{{}\begin{matrix}\dfrac{y}{x}+\dfrac{x}{y}\ge2\sqrt{\dfrac{y}{x}.\dfrac{x}{y}}=2\left(1\right)\\\dfrac{z}{x}+\dfrac{x}{z}\ge2\sqrt{\dfrac{z}{x}.\dfrac{x}{z}}=2\\\dfrac{z}{y}+\dfrac{y}{z}\ge2\sqrt{\dfrac{z}{y}.\dfrac{y}{z}}=2\left(3\right)\end{matrix}\right.\left(2\right)\)
Cộng từng vế 3 bất đẳng thức (1) (2) (3) , ta được:
A \(\ge6\)
Vậy \(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\ge6.\)Dấu '=' xảy ra khi a = b =c.
\(\dfrac{2a}{b+c-a}+\dfrac{2b}{c+a-b}+\dfrac{2c}{a+b-c}\)
\(=\dfrac{2a^2}{ab+ac-a^2}+\dfrac{2b^2}{ba+bc-b^2}+\dfrac{2c^2}{ca+cb-c^2}\)
\(\ge\dfrac{2\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)-a^2-b^2-c^2}\)
\(\ge\dfrac{2\left(a+b+c\right)^2}{\dfrac{\left(a+b+c\right)^2}{3}+a^2+b^2+c^2-a^2-b^2-c^2}=6\)
Dấu = xảy ra khi a = b = c
Áp dụng BĐT Bunhiacốpxki dạng phân thức có
\(\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge\dfrac{\left(a+b+c\right)^2}{a+2b^2+b+2c^2+c+2a^2}=\dfrac{9}{3+2\left(a^2+b^2+c^2\right)}\) (1)
Áp dụng BĐT Bunhiacốpxki có:
\(\left(a.1+b.1+c.1\right)^2\ge\left(1+1+1\right)\left(a^2+b^2+c^2\right)\)
\(\Rightarrow9\ge3\left(a^2+b^2+c^2\right)\Rightarrow3\ge a^2+b^2+c^2\Rightarrow2\left(a^2+b^2+c^2\right)\le6\) (2)
Thay (2) vào (1) có \(\dfrac{a^2}{a+b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+a^2}\ge\dfrac{9}{3+6}=1\) (đpcm)
Dấu = xảy ra khi a= b=c=1
Hình như sai đề :
Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\)
\(\Leftrightarrow\dfrac{ab+ac+bc}{abc}=0\)
\(\Leftrightarrow ab+ac+bc=0\) ( do \(a;b;c\ne0\) ) ( 1 )
Từ ( 1 ) \(\Rightarrow ab+bc=-ac\)
\(\Rightarrow\left(ab+bc\right)^2=\left[-\left(ac\right)\right]^2\)
\(\Rightarrow a^2b^2+b^2c^2+2ab^2c=a^2c^2\) ( * )
CMTT , ta được : \(\left\{{}\begin{matrix}b^2c^2+c^2a^2+2bc^2a=a^2b^2\\c^2a^2+a^2b^2+2a^2cb=b^2c^2\end{matrix}\right.\) ( *' )
Thay ( * ) và ( * ') vào E , ta được :
\(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-\left(a^2b^2+b^2c^2+2b^2ac\right)}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-\left(b^2c^2+c^2a^2+2bc^2a\right)}\)
\(+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-\left(c^2a^2+a^2b^2+2a^2cb\right)}\)
\(=\dfrac{a^2b^2c^2}{-2b^2ac}+\dfrac{a^2b^2c^2}{-2c^2ab}+\dfrac{a^2b^2c^2}{-2a^2cb}\)
\(=\dfrac{-ac}{2}+\dfrac{-ab}{2}+\dfrac{-bc}{2}\)
\(=\dfrac{-\left(ac+ab+bc\right)}{2}\)
\(=\dfrac{0}{2}=0\)
Vậy \(E=0\)
Lời giải:
\((3a+2b)(3a+2c)=16bc\)
\(\Leftrightarrow 9a^2+6a(b+c)=12bc\)
Theo BĐT Cô-si \(4bc\leq (b+c)^2\Rightarrow 9a^2+6a(b+c)\leq 3(b+c)^2\)
\(\Rightarrow 3a^2+2a(b+c)\leq (b+c)^2\)
\(\Leftrightarrow (b+c)^2-3a^2-2a(b+c)\geq 0\)
\(\Leftrightarrow (b+c)^2-9a^2-2a(b+c)+6a^2\geq 0\)
\(\Leftrightarrow (b+c-3a)(b+c+3a)-2a(b+c-3a)\geq 0\)
\(\Leftrightarrow (b+c-3a)(b+c+a)\geq 0\)
Vì $a+b+c>0$ nên \(b+c-3a\geq 0\Rightarrow b+c\geq 3a\) (đpcm)
b) Áp dụng BĐT Cô-si và kết quả phần a:
\(\frac{a}{b+c}+\frac{b+c}{a}=\frac{a}{b+c}+\frac{b+c}{9a}+\frac{8(b+c)}{9a}\)
\(\geq 2\sqrt{\frac{a}{b+c}.\frac{b+c}{9a}}+\frac{8(b+c)}{9a}=\frac{2}{3}+\frac{8(b+c)}{9a}\geq \frac{2}{3}+\frac{8.3a}{9a}=\frac{2}{3}+\frac{8}{3}=\frac{10}{3}\)
Ta có đpcm.
Ta có:
a/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a+2c}{3b+2d}\)
b/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-2a}{-2b}=\dfrac{7c}{7d}=\dfrac{-2a+7c}{-2b+7d}\)
PS: Xong
Áp dụng BĐT Cauchy cho 3 số dương a , b , c , ta có :
\(D=\dfrac{a}{a+2b}+\dfrac{b}{b+2c}+\dfrac{c}{c+2a}=\dfrac{a^2}{a^2+2ab}+\dfrac{b^2}{b^2+2bc}+\dfrac{c^2}{c^2+2ac}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)