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Ta có \(1+bc\ge abc+1\)
=>\(\frac{a}{1+bc}\le\frac{a}{abc+1}\)
Tương tự, + hết vào, ta có
\(P\le\frac{a+b+c}{abc+1}\)
Mà a,b,c\(\in\left[0;1\right]\Rightarrow\left(1-a\right)\left(1-b\right)+\left(1-c\right)\left(1-ab\right)\ge0\)
=>\(a+b+c\le abc+2\le2abc+2\Rightarrow\frac{a+b+c}{abc+1}\le2\) ( cái này nhân tung ra và rút gọn và có là abc >=0)
=> P<=2
dấu = xảy ra <=> 2 số = 1 và 1 số = 0
^_^
Ta có:
\(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow abc^2+ab^2c+a^2bc-ab-bc-ca=0\left(1\right)\)
Ta cần chứng minh
\(b\left(a^2-bc\right)\left(1-ac\right)=a\left(1-bc\right)\left(b^2-ac\right)\)
\(\Leftrightarrow ab^2c^2-a^2bc^2+ab^3c-b^2c-a^3bc+a^2c-ab^2+a^2b=0\)
\(\Leftrightarrow b\left(abc^2+ab^2c-bc-ab\right)-a^2bc^2-a^3bc+a^2c+a^2b=0\)
\(\Leftrightarrow b\left(ac-a^2bc\right)-a^2bc^2-a^3bc+a^2c+a^2b=0\)
\(\Leftrightarrow-a\left(ab^2c+abc^2+a^2bc-bc-ac-ab\right)=0\)(theo (1) thì đúng)
\(\RightarrowĐPCM\)
Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có:
\(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\)
\(\ge\left(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\) \(\left(1\right)\)
Lại có: \(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ac+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\) ( Do abc=1 )
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=1\) \(\left(2\right)\)
Từ (1) và (2) suy ra \(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\ge1\)
Mà \(a;b;c>0\Rightarrow a+b+c>0\)
\(\Rightarrow\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\ge\frac{1}{a+b+c}\) (đpcm)
\(a,b,c>0;ab+ac+bc=abc\)
<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z>0\)=> x + y + z = 1
Ta có:\(P=\frac{1}{bc\left(1+\frac{1}{a}\right)}+\frac{1}{ac\left(1+\frac{1}{b}\right)}+\frac{1}{ab\left(1+\frac{1}{c}\right)}\)
Viết lại \(P=\frac{yz}{1+x}+\frac{xz}{1+y}+\frac{xy}{1+z}\)
\(=\frac{yz}{\left(x+z\right)+\left(x+y\right)}+\frac{xz}{\left(x+y\right)+\left(z+y\right)}+\frac{xy}{\left(x+z\right)+\left(y+z\right)}\)
\(\le\frac{1}{4}\left(\frac{yz}{x+z}+\frac{yz}{x+y}\right)+\frac{1}{4}\left(\frac{xz}{x+y}+\frac{xz}{y+z}\right)+\frac{1}{4}\left(\frac{xy}{x+z}+\frac{xy}{y+z}\right)\)
\(\le\frac{1}{4}\left(\frac{yz+xy}{x+z}+\frac{yz+xz}{x+y}+\frac{xz+xy}{y+z}\right)=\frac{1}{4}\left(x+y+z\right)=\frac{1}{4}\)
Dấu "=" xảy ra <=> x = y = z = 1/3 <=> a= b = c = 3
max P = 1/4 tại a = b = c = 3
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)=\left(a+b\right)\left(a-b\right)+c\left(a-b\right)=\)\(\left(a-b\right)\left(a+b+c\right)\)
Tương tự:
\(b^2+ab-c^2-ac=\left(b-c\right)\left(a+b+c\right)\)
\(c^2+bc-a^2-ab=\left(c-a\right)\left(a+b+c\right)\)
\(Q=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)