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\(A=\dfrac{\left(a+b+c+a\right)\left(a+b+c+b\right)\left(a+b+c+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(A\ge\dfrac{2\sqrt{\left(a+b\right)\left(a+c\right)}.2\sqrt{\left(a+b\right)\left(b+c\right)}.2\sqrt{\left(a+c\right)\left(b+c\right)}}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=8\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(\frac{1}{a^4\left(1+b\right)\left(1+c\right)}=\frac{1}{\frac{a^4\left(1+b\right)\left(1+c\right)}{abc}}=\frac{\frac{1}{a^3}}{\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right)}\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\), tương tự suy ra:
\(A=\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+x\right)\left(1+z\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\)
Theo BĐT AM-GM ta có: \(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge\frac{3x}{4}\)
Tương tự suy ra \(A+\frac{3}{4}+\frac{x+y+z}{4}\ge\frac{3\left(x+y+z\right)}{4}\)
\(\Rightarrow A\ge\frac{x+y+z}{2}-\frac{3}{4}\ge\frac{3\sqrt[3]{xyz}}{2}-\frac{3}{4}=\frac{3}{4}\)
Dấu = xảy ra khi x=y=z=1 hay a=b=c=1
Cho e làm thử ạ:(
\(P=\frac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\)
\(=\frac{a+b+c+ab+bc+ca+abc+1}{1-\left(a+b+c\right)+ab+bc+ca-abc}\)
\(=1+\frac{2\left(a+b+c\right)+2abc}{1-\left(a+b+c\right)+\left(ab+bc+ca\right)-abc}\)
\(=1+\frac{2+2abc}{ab+bc+ca-abc}\)
Đặt \(\left(a+b+c;ab+bc+ca;abc\right)\rightarrow\left(p,q,r\right)\)
Khi đó \(P=1+\frac{2+2r}{q-r}\)
Áp dụng \(3q\le p^2\Rightarrow q\le\frac{1}{3}\Rightarrow P\ge1+\frac{2+2r}{\frac{1}{3}-r}=1+\frac{6+6r}{1-3r}\)
Sau khi đưa P về 1 biến thì e tịt ngòi r ạ:( Đến đây thì đi kiểu nào cx ngược dấu:(
Ta có: \(a+b+c=1\); a, b , c > 0 => 0 < a; b; c <1
=> \(\hept{\begin{cases}1+a=\left(1-b\right)+\left(1-c\right)\ge2\sqrt{\left(1-b\right)\left(1-c\right)}\\1+b=\left(1-c\right)+\left(1-a\right)\ge2\sqrt{\left(1-c\right)\left(1-a\right)}\\1+c=\left(1-a\right)+\left(1-b\right)\ge2\sqrt{\left(1-a\right)\left(1-b\right)}\end{cases}}\)
=> \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge8\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
=> \(P\ge8\)
"=" xảy ra <=> a = b =c = 1/ 3
\(A=\frac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\).
Ta có:
\(1-a=a+b+c-a\).(vì \(a+b+c=1\)).
\(\Leftrightarrow1-a=b+c\).
Chứng minh tương tự, ta được:
\(1-b=c+a\); \(1-c=a+b\). Do đó:
\(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(b+c\right)\left(c+a\right)\left(a+b\right)\).
Lại có:
\(1+a=a+b+c+a\)(vì \(a+b+c=1\)).
\(\Leftrightarrow1+a=\left(a+b\right)+\left(a+c\right)\).
Chứng minh tương tự, ta được:
\(1+b=\left(a+b\right)+\left(b+c\right)\); \(1+c=\left(a+c\right)+\left(b+c\right)\),.
Do đó \(\left(1+a\right)\left(1+b\right)\left(1+c\right)=\left[\left(a+b\right)+\left(a+c\right)\right]\left[\left(a+b\right)+\left(b+c\right)\right]\left[\left(a+c\right)+\left(b+c\right)\right]\)
Lúc đó:
\(A=\frac{\left[\left(a+b\right)+\left(a+c\right)\right]\left[\left(a+b\right)+\left(b+c\right)\right]\left[\left(a+c\right)+\left(b+c\right)\right]}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\).
Đặt \(a+b=x,b+c=y,c+a=z\left(x,y,z>0\right)\) thì \(x+y+z=2\left(a+b+c\right)=2\). Lúc đó:
\(A=\frac{\left(x+z\right)\left(x+y\right)\left(z+y\right)}{yzx}\).
Vì \(x,y>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:
\(x+z\ge2\sqrt{xz}\left(1\right)\).
Chứng minh tương tự, ta được:
\(x+y\ge2\sqrt{xy}\left(2\right)\);
\(z+y\ge2\sqrt{zy}\left(3\right)\).
Từ (1), (2), (3), ta được:
\(\left(x+z\right)\left(x+y\right)\left(z+y\right)\ge8\sqrt{xy.yz.zx}=8xyz\).
\(\Rightarrow\frac{\left(x+z\right)\left(x+y\right)\left(z+y\right)}{yzx}\ge\frac{8xyz}{xyz}=8\).
\(\Rightarrow A\ge8\).
Dấu bằng xảy ra.
\(\Leftrightarrow x=y=z>0\Leftrightarrow a+b=b+c=c+a>0\Leftrightarrow a=b=c>0\).
Mà \(a+b+c=1\)nên \(a=b=c=\frac{1}{3}\).
Vậy \(minA=8\Leftrightarrow a=b=c=\frac{1}{3}\).