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 kakakkakak tk cho bố m à

Bài 1 : Đề cần có điều kiện a,b,c là các số thực dương

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(=1\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)( vì \(a+b+c=1\))

Áp dụng BĐT Cauchy cho bộ 3 số dương ta có :

\(\hept{\begin{cases}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\end{cases}}\)

Khi đó : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot\frac{3}{\sqrt[3]{abc}}=\frac{3\cdot3\cdot\sqrt[3]{abc}}{\sqrt[3]{abc}}=9\)

Hay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)(đpcm)

Bài 2:

\(\hept{\begin{cases}b+c-a=x\\a+c-b=y\\a+b-c=z\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{y+z}{2}\\b=\frac{x+z}{2}\\c=\frac{x+y}{2}\end{cases}}}\)

\(\Rightarrow\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\)

\(=\frac{\frac{y+z}{2}}{x}+\frac{\frac{x+z}{2}}{y}+\frac{\frac{x+y}{2}}{z}\)

\(=\frac{y}{2x}+\frac{z}{2x}+\frac{x}{2y}+\frac{z}{2y}+\frac{x}{2z}+\frac{y}{2z}\)

Dùng AM-GM tự làm nốt

\(\hept{\begin{cases}b+c-a=x\\a+c-b=y\\a+b-c=z\end{cases}\Rightarrow}\hept{\begin{cases}a=\frac{z+y}{2}\\b=\frac{x+z}{2}\\c=\frac{y+x}{2}\end{cases}}\)

\(\Rightarrow\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}=\frac{y+z}{2x}+\frac{x+z}{2y}+\frac{y+x}{2z}=\frac{y}{2x}+\frac{z}{2x}+\frac{x}{2y}+\frac{z}{2y}+\frac{y}{2z}+\frac{x}{2z}\)Áp dụng BĐT AM-GM ta có:

\(\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge2.\sqrt{\frac{y}{2x}.\frac{x}{2y}}+2.\sqrt{\frac{z}{2x}.\frac{x}{2z}}+2.\sqrt{\frac{y}{2z}.\frac{z}{2y}}=1+1+1=3\)

Dấu " = " xảy ra <=> a=b=c

\(\frac{a}{b+c}>\frac{a}{a+b+c};\frac{b}{c+a}>\frac{b}{a+b+c};\frac{c}{a+b}>\frac{c}{a+b+c}\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)(1)

bạn tự c/m: \(\frac{a}{b}< \frac{a+c}{b+c}\left(b>a>0;c>0\right)\)

\(\Rightarrow\frac{a}{b+c}>\frac{2a}{a+b+c};\frac{b}{c+a}< \frac{2b}{a+b+c};\frac{c}{a+b}< \frac{2c}{a+b+c}\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< \frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)(2)

Từ (1) và (2) 

\(1< \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< 2\)

đpcm

https://olm.vn/hoi-dap/detail/12121415915.html

vô đi rồi k cho mk

Ta co:

\(\frac{a^2}{ab+ca-a^2}+\frac{b^2}{ab+bc-b^2}+\frac{c^2}{ca+bc-c^2}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)}\)

\(\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca}\ge\frac{\left(a+b+c\right)^2}{\frac{\left(a+b+c\right)^2}{3}}=3\)

Dau '=' xay ra khi \(a=b=c\)

Ta có :\(2\sqrt{\frac{b+c-a}{a}}\le\frac{b+c-a}{a}+1=\frac{b+c}{a}\)

<=>   \(\sqrt{\frac{a}{b+c-a}}\ge\frac{2a}{b+c}\)

\(CMTT\)=> \(\sqrt{\frac{b}{c+a-b}}\ge\frac{2b}{c+a}\)

                      \(\sqrt{\frac{c}{a+b-c}}\ge\frac{2c}{a+b}\)

=>\(VT\)\(\ge\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}\)

\(CM\)\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)

=>    \(\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}\ge3\)

=>\(VT\ge3\)

lên google tìm cosi mà làm theo nha

\(\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\)

\(=\frac{a^2}{ab+ca-a^2}+\frac{b^2}{ab+bc-b^2}+\frac{c^2}{ca+bc-c^2}\)

\(\ge\frac{\left(a+b+c\right)^2}{2ab+2bc+2ca-a^2-b^2-c^2}\)

\(\ge\frac{3\left(ab+bc+ca\right)}{2ab+2bc+2ca-ab-bc-ca}=3\)

\(VT=\frac{2\left(a-b\right)^2}{\left(b+c-a\right)\left(c+a-b\right)}+\frac{2\left(b-c\right)^2}{\left(c+a-b\right)\left(a+b-c\right)}+\frac{2\left(a-c\right)^2}{\left(a+b-c\right)\left(b+c-a\right)}+3\ge3\)

Cách khác Đặt \(\hept{\begin{cases}b+c-a=x\\a+c-b=y\\a+b-c=z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}c=\frac{x+y}{2}\\b=\frac{x+z}{2}\\a=\frac{z+y}{2}\end{cases}}\)

BĐT cần CM \(\Leftrightarrow\frac{z+y}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}\ge3\)

\(\Leftrightarrow\frac{z}{x}+\frac{y}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\ge6\)

Đúng (do \(\frac{x}{y}+\frac{y}{x}\ge2;\frac{x}{z}+\frac{z}{x}\ge2;\frac{z}{y}+\frac{y}{z}\ge2\))

Dấu "="

b)\(\Sigma\frac{a}{b+c-a}=\Sigma\frac{a^2}{ab+bc-a^2}\)\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)}\)(BĐT Svarxơ)\(\ge\frac{\left(a+b+c\right)^2}{\frac{2}{3}\left(a+b+c\right)^2-\frac{1}{3}\left(a+b+c\right)^2}\)(BĐT Bunhiacopxki)\(=3\)(đpcm)

a)\(\Sigma\frac{a}{b+c}=\Sigma\frac{a^2}{ab+bc}\)\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)\(\ge\frac{\left(a+b+c\right)^2}{\frac{2}{3}\left(a+b+c\right)^2}=1,5>1\)

Do a;b;c là 3 cạnh tam giác nên

\(\hept{\begin{cases}a+b>c\\b+c>a\\c+a>b\end{cases}\Rightarrow\hept{\begin{cases}a+b-c>0\\b+c-a>0\\c+a-b>0\end{cases}}}\)

Đặt \(b+c-a=x;a+c-b=y;a+b-c=z\)

Gọi \(A=\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\)

\(\Rightarrow2A=\frac{\left(y+z\right)}{x}+\frac{\left(x+z\right)}{y}+\frac{\left(x+y\right)}{z}\)

            \(=\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\)

Rồi dùng Cô-si

\(\Rightarrow2A\ge6\)

\(\Leftrightarrow A\ge3\)

Dấu = xảy ra khi a=b=c

 Áp dụng BĐT AM-GM ta có \(\text{∑}\frac{3}{b+c-a}\ge3\sqrt[3]{\frac{abc}{\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}}\ge3\)

Dấu đẳng thức xảy ra khi và chỉ khi a = b = c. 

\(A=\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\)

\(=\frac{a^2}{ab+ac-a^2}+\frac{b^2}{ba+bc-b^2}+\frac{c^2}{ca+cb-c^2}\)

\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)}\)

\(\ge\frac{\left(a+b+c\right)^2}{\frac{2\left(a+b+c\right)^2}{3}-\frac{\left(a+b+c\right)^2}{3}}=3\)