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dễ mà
\(A=\frac{abc}{c^3}+\frac{abc}{b^3}+\frac{abc}{a^3}\)
\(A=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)\(\Rightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
ta có
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\left(\frac{1}{a}+\frac{1}{b}\right)^3+\frac{1}{c^3}-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}-\frac{1}{ac}-\frac{1}{bc}\right)+\frac{3}{abc}=\frac{3}{abc}\)
\(A=abc.\frac{3}{abc}=3\)
Chú ý rằng nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz.\).
Nếu vậy : \(x+y+z=0\) \(\Rightarrow z=-\left(x+y\right).\)
Do đó : \(x^3+y^3+z^3=x^3+y^3-\left(x+y\right)^3=-3x^2y-3xy^2=-3xy\left(x+y\right)=3xyz\)
Từ đây có thể suy ra :
\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
Áp dụng nhận xét trên,ta có :
Nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) thì \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}=\frac{3}{abc}.\)
Do đó : \(N=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\) với \(a,b,c\ne0\).
Đặt \(A=abc\left(bc+a^2\right)\left(ac+b^2\right)\left(ab+c^2\right)\)
Do a; b; c > 0 => A > 0
Giả sử \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a+b}{bc+a^2}-\frac{b+c}{ac+b^2}-\frac{c+a}{ab+c^2}\ge0\)
\(\Leftrightarrow\frac{a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-b^4a^2c^2-c^4a^2b^2}{A}\ge0\)( tự quy đồng rồi rút gọn nhé, làm chi tiết dài lắm )
\(\Leftrightarrow\frac{2a^4b^4+2b^4c^4+2c^4a^4-2a^4b^2c^2-2b^4a^2c^2-2c^4a^2b^2}{A}\ge0\)
\(\Leftrightarrow\frac{\left(a^2b^2+b^2c^2\right)^2+\left(b^2c^2+c^2a^2\right)^2+\left(c^2a^2+a^2b^2\right)^2}{A}\ge0\)(đúng)
Vậy \(\frac{a+b}{bc+a^2}+\frac{b+c}{ca+b^2}+\frac{c+a}{ab+c^2}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)(đpcm)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Rightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-ac-bc\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)
\(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự: \(b^2+2ac=\left(b-c\right)\left(b-a\right)\)
\(c^2+2ab=\left(a-c\right)\left(b-c\right)\)
\(B=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}+\frac{ca+1}{\left(b-a\right)\left(b-c\right)}+\frac{ab+1}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}-\frac{ca+1}{\left(a-b\right)\left(b-c\right)}+\frac{ab+1}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-c\right)+\left(ab+1\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-c\right)+\left(ab+1\right)\left(a-b\right)\)
\(=\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-b\right)-\left(ca+1\right)\left(b-c\right)+\left(ab+1\right)\left(a-b\right)\)
\(=\left(b-c\right)\left(bc+1-ca-1\right)+\left(a-b\right)\left(ab+1-ca-1\right)\)
\(=\left(b-c\right)\left(bc-ca\right)+\left(a-b\right)\left(ab-ca\right)\)
\(=\left(b-c\right)c\left(b-a\right)+\left(a-b\right)a\left(b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
Vậy B = 1
Câu 1:
- Chứng minh a3+b3+c3=3abc thì a+b+c=0
\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3abc\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow0=0\) Đúng (Đpcm)
- Chứng minh a3+b3+c3=3abc thì a=b=c
Áp dụng Bđt Cô si 3 số ta có:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu = khi a=b=c (Đpcm)
Câu 2
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\cdot\frac{1}{abc}\)
Ta có:
\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\cdot3\cdot\frac{1}{abc}=3\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=>\(\frac{1}{a}=-\left(\frac{1}{b}+\frac{1}{c}\right)\)
=>\(\frac{1}{a^2}=-\left(\frac{1}{ab}+\frac{1}{ca}\right)\)
cm tương tự: \(\frac{1}{b^2}=-\left(\frac{1}{ab}+\frac{1}{bc}\right)\)
\(\frac{1}{c^2}=-\left(\frac{1}{ca}+\frac{1}{bc}\right)\)
=> \(N=-\left[bc\left(\frac{1}{ab}+\frac{1}{ca}\right)+ca\left(\frac{1}{ab}+\frac{1}{bc}\right)+ab\left(\frac{1}{ca}+\frac{1}{bc}\right)\right]\)
\(=-\left[\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right]\)
\(=-\left[\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right]\) (1)
Ta có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=>\(\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}=0\)
=>\(1+\frac{b+c}{a}+1+\frac{a+c}{b}+1+\frac{a+b}{c}=0\)
=>\(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=-3\) (2)
Từ (1) và (2) =>N=3
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
Làm tương tự. có: \(\frac{b^2+ca}{b^2+2ca}=\frac{b^2+ca}{b^2+ca-ab-bc}=\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}\)
\(\frac{c^2+ab}{c^2+2ab}=\frac{c^2+ab}{c^2+ab-ca-bc}=\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(\Rightarrow A=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}+\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}+\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(=\frac{\left(a^2+bc\right).\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{\left(b^2+ca\right).\left(a-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}+\frac{\left(c^2+ab\right).\left(a-b\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
Sau đó bạn thực hiện tiếp nhé.
Bài 1: Cho \(a,b,c\ge0:a^2+b^2+c^2=3\). CMR: \(a^4b^4+b^4c^4+c^4a^4\le3\)
Bài 2: Cho \(a,b,c\ge0\). CMR: \(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)
Bài 3: Cho \(a,b,c\ge0:a^2+b^2+c^2=a+b+c\). CMR: \(a^2b^2+b^2c^2+c^2a^2\le ab+bc+ca\)
Bài 4: Cho \(a,b,c\ge0\). CMR: \(4\left(a+b+c\right)^3\ge27\left(ab^2+bc^2+ca^2+abc\right)\)
Bài 5: Cho \(a,b,c\ge0:a+b+c=3\).CMR: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+3.\frac{1}{a^2}.\frac{1}{b}+3.\frac{1}{a}.\frac{1}{b^2}+\frac{1}{b^3}=-\frac{1}{c}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{a^2b}+\frac{3}{ab^2}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}.\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}.\left(-\frac{1}{c}\right)=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}-\frac{3}{abc}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Do đó:
\(A=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc.\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)