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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ac}{abc}=0\Rightarrow ab+bc+ac=0\Rightarrow bc=-ab-ac\)
=> \(\frac{1}{a^2+2bc}=\frac{1}{a^2+bc-ab-ac}=\frac{1}{a\left(a-b\right)-c\left(a-b\right)}=\frac{1}{\left(a-b\right)\left(a-c\right)}\)tương tự thì
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(a-b\right)\left(c-b\right)}+\frac{1}{\left(c-b\right)\left(c-a\right)}=\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{c-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{bc+ca+ab}{abc}=0\)
\(\Leftrightarrow bc+ca+ab=0\)
\(\Leftrightarrow\hept{\begin{cases}bc=-ab-ca\\ca=-ab-bc\\ab=-ca-bc\end{cases}}\)
Ta có : \(A=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(\Leftrightarrow A=\frac{a^2}{a^2+bc-ab-ca}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-ca-bc}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a+b\right)\left(a-b\right)\left(b-c\right)-\left(b+c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(b-c\right)\left[\left(a+b\right)-\left(b+c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
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: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)
Ta có:
\(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) ( luôn đúng )
Áp dụng:
\(G=\frac{a^3+b^3}{2ab}+\frac{b^3+c^3}{2bc}+\frac{c^3+a^3}{2ca}\)
\(\ge\frac{ab\left(a+b\right)}{2ab}+\frac{bc\left(b+c\right)}{2bc}+\frac{ca\left(c+a\right)}{2ca}\)
\(=\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}\)
\(=a+b+c=2019\)
Dấu "=" xảy ra tại a=b=c=673
Hình như đề là a2+b2 thôi chứ có cả 1+a2+b2 luôn à? Mình làm theo cái đề có a2+b2 chứ không có +1 nhé!
Áp dụng bất đẳng thức Cauchy Schawrz dạng Engel ta được:
\(B=\frac{1^2}{a^2+b^2}+\frac{1^2}{2ab}\ge\frac{\left(1+1\right)^2}{a^2+b^2+2ab}=\frac{4}{\left(a+b\right)^2}\)
mà a;b>0 => a+b>0 và \(a+b\le1\Rightarrow\left(a+b\right)^2\le1\) => \(\frac{4}{\left(a+b\right)^2}\ge\frac{4}{1}=4\)
=>\(B=\frac{1}{a^2+b^2}+\frac{1}{2ab}\ge4\Rightarrow B_{min}=4\) <=> a=b=0,5
@Trà My: có 1+a2+b2 thì vẫn có Min vấn đề là chưa đủ trình độ mà còn đòi tự sửa đề