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Theo BĐT Schur thì ta có:
\((a+b-c)(b+c-a)(c+a-b)\leq abc\)
Vậy thì giờ chỉ theo AM-GM là xong
\(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
\(\ge3\sqrt[3]{\dfrac{abc}{\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}}=3\)
Ta có : Do a ; b ; c là 3 cạnh của 1 tam giác nên :
\(\dfrac{a}{a+b+c}< \dfrac{a}{b+c}< \dfrac{2a}{a+b+c}\)
\(\dfrac{b}{a+b+c}< \dfrac{b}{c+a}< \dfrac{2b}{a+b+c}\)
\(\dfrac{c}{a+b+c}< \dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\)
Cộng 3 vế với nhau , ta có :
\(1< \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\left(đpcm\right)\)
Ta có :
\(\dfrac{â}{b+c}>\dfrac{a}{a+b+c}\);
\(\dfrac{b}{c+a}>\dfrac{b}{a+b+c}\);
\(\dfrac{c}{a+b}>\dfrac{c}{a+b+c}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}>\dfrac{a+b+c}{a+b+c}=1\) (*)
Ta có bất đằng thức tam giác : a+b > c ; b+c > a ; a+c > b
\(\Rightarrow\dfrac{a}{b+c}< 1;\dfrac{b}{a+c}< 1;\dfrac{c}{a+b}< 1\)
Vì \(\dfrac{a}{b+c}< 1\Rightarrow\dfrac{a}{b+c}< \dfrac{a+a}{a+b+c}=\dfrac{2a}{a+b+c}\)
Tương tự :
\(\dfrac{b}{a+c}< \dfrac{2b}{a+b+c};\dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}< \dfrac{2\left(a+b+c\right)}{a+b+c}=2\) (**)
Kết hợp (*) với (**)
=> ĐPCM
-Đặt \(\left\{{}\begin{matrix}b+c-a=x>0\\c+a-b=y>0\\a+b-c=z>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2c=x+y\\2a=y+z\\2b=z+x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2}\\b=\dfrac{z+x}{2}\end{matrix}\right.\)
\(A=\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}=\dfrac{\dfrac{y+z}{2}}{x}+\dfrac{\dfrac{z+x}{2}}{y}+\dfrac{\dfrac{x+y}{2}}{z}=\dfrac{1}{2}\left(\dfrac{y+z}{x}+\dfrac{z+x}{y}+\dfrac{x+y}{z}\right)=\dfrac{1}{2}\left[\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)+\left(\dfrac{x}{z}+\dfrac{z}{x}\right)\right]\ge\dfrac{1}{2}.\left(2+2+2\right)=3\left(đpcm\right)\)
-Dấu "=" xảy ra khi \(a=b=c\)
Đặt: \(b+c-a=x\)
\(a+c-b=y\)
\(a+b-c=z\)
Suy ra:
\(2a=y+z\)
\(2b=x+z\)
\(2c=x+y\)
Ta có:
\(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)
\(=\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\) ( BĐT luôn đúng)
=> ĐPCM
a,b,c là độ dài 3 cạnh t/g
\(\Rightarrow\dfrac{a}{b+c-a};\dfrac{b}{a+c-b};\dfrac{c}{a+b-c}>0\)
\(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
\(A+\dfrac{3}{2}=\dfrac{a}{b+c-a}+\dfrac{1}{2}+\dfrac{b}{a+c-b}+\dfrac{1}{2}+\dfrac{c}{b+a-c}+\dfrac{1}{2}\)
\(A+\dfrac{3}{2}=\dfrac{a+b+c}{2\left(b+c-a\right)}+\dfrac{a+b+c}{2\left(a+c-b\right)}+\dfrac{a+b+c}{2\left(b+a-c\right)}\)
\(A+\dfrac{3}{2}=\dfrac{\left(a+b+c\right)}{2}\left(\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}+\dfrac{1}{b+a-c}\right)\)
\(A+\dfrac{3}{2}\ge\dfrac{a+b+c}{2}\cdot\dfrac{9}{b+c-a+a+c-b+b+a-c}\)
\(A+\dfrac{3}{2}\ge\dfrac{9}{2}\)
\(\Rightarrow A\ge3\left(đpcm\right)\)
a)a,b,c là độ dài 3 cạnh của 1 tam giác
\(\Rightarrow a< b+c\Rightarrow a^2< ab+ac\)
TT\(\Rightarrow b^2< ba+bc\)
\(c^2< ca+cb\)
Cộng vế theo vế ta có đpcm
b)BĐT\(\Leftrightarrow\dfrac{a}{b+c-a}+\dfrac{1}{2}+\dfrac{b}{a+c-b}+\dfrac{1}{2}+\dfrac{c}{a+b-c}+\dfrac{1}{2}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{a+b+c}{b+c-a}+\dfrac{a+b+c}{a+c-b}+\dfrac{a+b+c}{a+b-c}\right)\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}+\dfrac{1}{a+b-c}\right)\ge9\)(đúng theo AM-GM)
câu 1 :Đặt b+c-a=x; a+c-b=y ; a+b-c=z
vì a,b,c là 3 cạnh của tam giác nên
b+c-a>0 ; a+c-b>0 ; a+b-c>0
Đặt biểu thức \(\dfrac{a}{b +c-a}\)+\(\dfrac{b}{c+a-b}\)+\(\dfrac{c}{a+b-c}\)=S thì
2S=\(\dfrac{2a}{b+c-a}\)+\(\dfrac{2b}{c+a-b}\)+\(\dfrac{2c}{a+b-c}\)
mà \(\dfrac{2a}{b+c-a}\)=\(\dfrac{a+c-b+a+b-c}{b+c-a}\)=\(\dfrac{y+z}{x}\) , tương tự
\(\dfrac{2b}{c+a-b}\)=\(\dfrac{x+z}{y}\)
\(\dfrac{2c}{a+b-c}\)=\(\dfrac{x+y}{z}\)
=>2S=\(\dfrac{x+y}{z}\)+\(\dfrac{y+z}{x}\)+\(\dfrac{x+z}{y}\)=\(\dfrac{x}{z}\)+\(\dfrac{y}{z}\)+\(\dfrac{y}{x}\)+\(\dfrac{z}{x}\)+\(\dfrac{x}{y}\)+\(\dfrac{z}{y}\)
ta thấy \(\dfrac{x}{z}\)+\(\dfrac{z}{x}\)=\(\dfrac{x^{2^{ }}+z^2}{xz}\)\(\ge\)\(\dfrac{2xz}{xz}\)=2 tương tự với 2 cặp số nghich đảo còn lại thì ta có 2S\(\ge\)2+2+2=6
nên S\(\ge\)3
dấu = xảy ra \(\Leftrightarrow\)x=y=z
câu 2 :
ta có a+b>c ;b+c>a ; a+c>b
xét \(\dfrac{1}{a+c}\)+\(\dfrac{1}{b+c}\)>\(\dfrac{1}{a+b+c}\)+\(\dfrac{1}{b+c+a}\)=\(\dfrac{2}{a+b+c}\)>\(\dfrac{2}{a+b+a+b}\)=\(\dfrac{1}{a+b}\)
tương tự \(\dfrac{1}{a+b}\)+\(\dfrac{1}{a+c}\)>\(\dfrac{1}{b+c}\);\(\dfrac{1}{a+b}\)+\(\dfrac{1}{b+c}\)>\(\dfrac{1}{a+c}\)
nên điều phải chứng minh
\(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\)
\(\Leftrightarrow\dfrac{a}{b+c-a}+\dfrac{1}{2}+\dfrac{b}{a+c-b}+\dfrac{1}{2}+\dfrac{c}{a+b-c}+\dfrac{1}{2}\ge3+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a+b+c}{2\left(b+c-a\right)}+\dfrac{a+b+c}{2\left(a+c-b\right)}+\dfrac{a+b+c}{2\left(a+b-c\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\dfrac{a+b+c}{2}\left(\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}+\dfrac{1}{a+b-c}\right)\ge\dfrac{9}{2}\)
Lại có:\(\dfrac{a+b+c}{2}\left(\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}+\dfrac{1}{a+b-c}\right)\ge\dfrac{a+b+c}{2}\cdot\dfrac{9}{b+c-a+a+c-b+a+b-c}\ge\dfrac{9}{2}\left(đpcm\right)\)
Lời giải:
Có nhiều cách để giải quyết bài toán này. Đây là một cách đơn thuần sử dụng BĐT Cô-si.
Đặt \(\left\{\begin{matrix} b+c-a=x\\ a+c-b=y\\ a+b-c=z\end{matrix}\right.\) (\(x,y,z>0\) do $a,b,c$ là ba cạnh tam giác)
\(\Rightarrow (a,b,c)=\left(\frac{y+z}{2}; \frac{x+z}{2}; \frac{x+y}{2}\right)\)
BĐT cần chứng minh tương đương với:
\(\frac{y+z}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}\geq 3(*)\)
Áp dụng BĐT Cô-si cho 3 số:
\(\frac{y+z}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}\geq 3\sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8xyz}}\)
Tiếp tục Cô-si: \((x+y)(y+z)(z+x)\geq 2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}=8xyz\)
\(\Rightarrow \frac{y+z}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}\geq 3\sqrt[3]{\frac{8xyz}{8xyz}}=3\)
Do đó $(*)$ được chứng minh.
Dấu bằng xảy ra khi \(x=y=z\Leftrightarrow a=b=c\)