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\(\frac{2a^2-2ac+c^2}{2b^2-2bc+c^2}=\frac{a-c}{b-c}\)
\(\Leftrightarrow2a^2b-2a^2c+ac^2-bc^2-2ab^2+2b^2c=0\)
\(\Leftrightarrow2a\left(ab-ac+\frac{c^2}{2}\right)-bc^2-2ab^2+2bc^2=b\left(2ac-c^2-2ab+2bc\right)=0\)(đúng)
=> đpcm
Từ \(c^2+2\left(ab-bc-ac\right)=0.\)
\(\Rightarrow c^2+2ab-2bc-2ac=0\)
\(\Rightarrow\frac{c^2}{2}+ab-bc-ac=0\)
\(\Rightarrow bc=\frac{c^2}{2}+ab-ac\)
Có : \(2a\left(ab-ac+\frac{c^2}{2}\right)-bc^2-2ab^2+2bc^2\)
\(=2abc-bc^2-2ab^2+2bc^2\)
\(=-b\left(-2ac+c^2+2ab-2bc\right)\)
\(=-b\left[c^2+2\left(ab-bc-ac\right)\right]=-b.0=0\)\(\left(đpcm\right)\)
\(\left\{{}\begin{matrix}c^2-2ca+a^2+2ab-2bc=a^2\\c^2-2bc+b^2+2ab-2ac=b^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a-c\right)^2+2b\left(a-c\right)=a^2\\\left(b-c\right)^2+2a\left(b-c\right)=b^2\end{matrix}\right.\)
\(\Rightarrow\frac{a^2+a^2-2ac+c^2}{b^2+b^2-2bc+c^2}=\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2}{\left(b-c\right)^2+2a\left(b-c\right)+\left(b-c\right)^2}\)
\(=\frac{2\left(a-c\right)^2+2b\left(a-c\right)}{2\left(b-c\right)^2+2a\left(b-c\right)}=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}=\frac{a-c}{b-c}\)
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
Do \(c^2+2\left(ab-ac-bc\right)=0\Leftrightarrow-c^2=2\left(ab-ac-bc\right)\)
Ta có; \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+c^2-c^2+\left(a-c\right)^2}{b^2+c^2-c^2+\left(b-c\right)^2}=\frac{a^2+c^2+2\left(ab-ac-bc\right)+\left(a-c\right)^2}{b^2+c^2+2\left(ab-ac-bc\right)+\left(b-c\right)^2}\)
\(=\frac{2\left(a-c\right)^2+2\left(ab-bc\right)}{2\left(b-c\right)^2+2\left(ab-ac\right)}=\frac{2\left(a-c\right)^2+2b\left(a-c\right)}{2\left(b-c\right)^2+2a\left(b-c\right)}=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}\)
\(=\frac{a-c}{b-c}\) (đpcm)