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Đặt \(A=\frac{c\left(ab+1\right)^2}{b^2\left(bc+1\right)}+\frac{a\left(bc+1\right)^2}{c^2\left(ca+1\right)}+\frac{b\left(ca+1\right)^2}{a^2\left(ab+1\right)}\) và \(x=ab+1;\) \(y=bc+1;\) \(z=ca+1\) \(\left(\text{*}\right)\)
Khi đó, với các giá trị tương ứng trên thì biểu thức \(A\) trở thành: \(A=\frac{cx^2}{b^2y}+\frac{ay^2}{c^2z}+\frac{bz^2}{a^2x}\)
Áp dụng bất đẳng thức Cauchy cho bộ ba phân số không âm của biểu thức trên (do \(a,b,c>0\)), ta có:
\(A=\frac{cx^2}{b^2y}+\frac{ay^2}{c^2z}+\frac{bz^2}{a^2x}\ge3\sqrt[3]{\frac{cx^2}{b^2y}.\frac{ay^2}{c^2z}.\frac{bz^2}{a^2z}}=3\sqrt[3]{\frac{xyz}{abc}}\) \(\left(\text{**}\right)\)
Mặt khác, do \(ab+1\ge2\sqrt{ab}\) (bất đẳng thức AM-GM cho hai số \(a,b\) luôn dương)
nên \(x\ge2\sqrt{ab}\) \(\left(1\right)\) (theo cách đặt ở \(\left(\text{*}\right)\))
Hoàn toàn tương tự với vòng hoán vị \(a\) \(\rightarrow\) \(b\) \(\rightarrow\) \(c\) và với chú ý cách đặt ở \(\left(\text{*}\right)\), ta cũng có:
\(y\ge2\sqrt{bc}\) \(\left(2\right)\) và \(z\ge2\sqrt{ca}\) \(\left(3\right)\)
Nhân từng vế \(\left(1\right);\) \(\left(2\right)\) và \(\left(3\right)\), ta được \(xyz\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc\)
Do đó, \(3\sqrt[3]{\frac{xyz}{abc}}\ge3\sqrt[3]{\frac{8abc}{abc}}=3\sqrt[3]{8}=6\) \(\left(\text{***}\right)\)
Từ \(\left(\text{**}\right)\) và \(\left(\text{***}\right)\) suy ra được \(A\ge6\), tức \(\frac{c\left(ab+1\right)^2}{b^2\left(bc+1\right)}+\frac{a\left(bc+1\right)^2}{c^2\left(ca+1\right)}+\frac{b\left(ca+1\right)^2}{a^2\left(ab+1\right)}\ge6\) (điều phải chứng minh)
Dấu \("="\) xảy ra \(\Leftrightarrow\) \(a=b=c=1\)
Lời giải :
\(A=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{-bc\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-ac\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=\frac{-bc\left(b-c\right)-ac\left(c-a\right)-ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Xét tử số :
\(TS=-b^2c+bc^2-ac^2+a^2c-a^2b+ab^2\)
\(=-ab\left(a-b\right)-c^2\left(a-b\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(-ab-c^2+ac+bc\right)\)
\(=\left(a-b\right)\left[-a\left(b-c\right)+c\left(b-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Khi đó \(A=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)=\left(a+b\right)\left(a-b\right)+c\left(a-b\right)=\)\(\left(a-b\right)\left(a+b+c\right)\)
Tương tự:
\(b^2+ab-c^2-ac=\left(b-c\right)\left(a+b+c\right)\)
\(c^2+bc-a^2-ab=\left(c-a\right)\left(a+b+c\right)\)
\(Q=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Vì \(c^2+2\left(ab-ac-bc\right)=0\) nên :
\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+\left(a-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}{b^2+\left(b-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}\)
\(=\frac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}=\frac{\left(a-c\right)^2+b\left(a-c\right)}{\left(b-c\right)^2+a\left(b-c\right)}\)
\(=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}=\frac{a-c}{b-c}\) \(\left(b\ne c,a+b\ne0\right)\)
Đặt \(A=\frac{\left(a+b\right)^2}{ab}+\frac{\left(b+c\right)^2}{bc}+\frac{\left(c+a\right)^2}{ca}=\frac{a^2+2ab+b^2}{ab}+\frac{b^2+2bc+c^2}{bc}+\frac{c^2+2ac+c^2}{ca}\)
\(=\frac{a}{b}+2+\frac{b}{a}+\frac{b}{c}+2+\frac{c}{b}+\frac{c}{a}+2+\frac{a}{c}=6+a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{b}+\frac{1}{a}\right)\)
\(\ge6+\frac{4a}{b+c}+\frac{4b}{c+a}+\frac{4c}{a+b}\ge6+2\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+b}\right)+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
\(\ge6+2\cdot\frac{3}{2}+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=9+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
Dấu "=" xảy ra <=> a=b=c
\(a,b,c>0;ab+ac+bc=abc\)
<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
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
ĐK: a,b,c>0
\(VT\ge2.2.2.\sqrt{a.\frac{b}{ac}}.\sqrt{b.\frac{c}{ab}}.\sqrt{c.\frac{a}{bc}}\)
\(=8.\sqrt{\frac{abc}{abc}}=8^{\left(đpcm\right)}\)
Dấu bằng xảy ra khi a = b =c