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Vì a+b+c=0=>(a+b)=-c. Tương tự:(b+c)=-a;(a+c)=-b.
Ta có A=:\(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\frac{b^2}{\left(b-c\right)\left(b+c\right)-a^2}+\frac{c^2}{\left(c-a\right)\left(c+a\right)-b^2}\)
\(=\frac{a^2}{\left(a-b\right).\left(-c\right)-c^2}+tươngtự\)
\(=\frac{a^2}{-ca+bc-c^2}\)+ tương tự
\(=\frac{a^2}{c\left(b-c-a\right)}+tươngtự\)
\(=\frac{a^2}{c\left(b-\left(c+a\right)\right)}\)+ tương tự nha
\(=\frac{a^2}{c\left(b-\left(-b\right)\right)}+tươngtự=\frac{a^2}{2bc}+tươngtự\)
Sau đó ta có :\(\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2bc}\)
=\(\frac{a^3+b^3+c^3}{2abc}=\frac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{2abc}\)
\(=\frac{\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b\right)}{2abc}\)=\(\frac{0-0-3ab\left(-c\right)}{2abc}\)(do a+b+c=0)
=\(\frac{3abc}{2abc}=\frac{3}{2}\)Ok r bạn
Ta có
\(\frac{a^2}{a+b^2}=\frac{a^2+ab^2-ab^2}{a+b^2}=a-\frac{ab^2}{a+b^2}\ge a-\frac{b\sqrt{a}}{2}\ge a-\frac{1}{4}b\left(a+1\right)\)
Khi đó
\(A\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}\left(ab+bc+ac\right)\)
Mà \(ab+bc+ac\le\frac{1}{3}\left(a+b+c\right)^2=3\)
=> \(A\ge\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\)( ĐPCM)
Dấu bằng xảy ra khi a=b=c=1
\(a-\frac{ab^2}{a+b^2}\ge a-\frac{b\sqrt{a}}{2}\)
Do \(a+b^2\ge2b\sqrt{a}\)
\(a-\frac{ab^2}{a+b^2}\ge a-\frac{b\sqrt{a}}{2}\ge a-\frac{1}{4}b\left(a+1\right)\)
Do \(\sqrt{a}\le\frac{a+1}{2}\)
Xét: \(9M=\Sigma\frac{a^2+b^2+c^2}{4a^2+b^2+c^2}-\frac{3}{2}+\Sigma\frac{2\left(ab+bc+ca\right)}{4a^2+b^2+c^2}-3+\frac{9}{2}\)
\(=\Sigma\left(\frac{a^2+b^2+c^2}{4a^2+b^2+c^2}-\frac{1}{2}\right)+\Sigma\left(\frac{2\left(ab+bc+ca\right)}{4a^2+b^2+c^2}-1\right)+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\frac{b^2+c^2-2a^2}{\left(4a^2+b^2+c^2\right)}+\Sigma\frac{2ab+2bc+2ca-4a^2-b^2-c^2}{4a^2+b^2+c^2}+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\frac{\left(b-a\right)\left(b+a\right)+\left(c-a\right)\left(c+a\right)}{\left(4a^2+b^2+c^2\right)}+\Sigma\frac{2a\left[\left(b-a\right)+\left(c-a\right)\right]}{4a^2+b^2+c^2}-\Sigma\frac{\left(b-c\right)^2}{4a^2+b^2+c^2}+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\left(\frac{\left(a-b\right)\left(a+b\right)}{a^2+4b^2+c^2}-\frac{\left(a-b\right)\left(b+a\right)}{4a^2+b^2+c^2}\right)-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}-\Sigma\frac{\left(a-b\right)^2}{a^2+b^2+4c^2}+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\left(a-b\right)\left(a+b\right)\left(\frac{3a^2-3b^2}{\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}\right)-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}-\Sigma\frac{\left(a-b\right)^2}{a^2+b^2+4c^2}+\frac{9}{2}\)
\(=\Sigma\frac{3\left(a-b\right)^2\left(a+b\right)^2}{2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}-\Sigma\frac{\left(a-b\right)^2}{a^2+b^2+4c^2}+\frac{9}{2}\)
\(=\Sigma\left(a-b\right)^2\left[\frac{3\left(a+b\right)^2}{2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\frac{1}{a^2+b^2+4c^2}\right]-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}+\frac{9}{2}\)
\(=\Sigma\left(a-b\right)^2\left[\frac{3\left(a+b\right)^2\left(a^2+b^2+4c^2\right)-2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}{2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(a^2+b^2+4c^2\right)}\right]-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}+\frac{9}{2}\)Ai đó làm tiếp giúp em vs:( Em chỉ nghĩ ra được tới đây thôi.
Ta có:
\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;a^2+c^2\ge2\sqrt{a^2c^2}=2ac;a^2+a^2\ge2\sqrt{a^2a^2}=2a^2\)
Khi đó:
\(4a^2+b^2+c^2\ge2a\left(a+b+c\right)\)
\(\Rightarrow\frac{1}{4a^2+b^2+c^2}\le\frac{1}{6a}\)
Tương tự:
\(\frac{1}{a^2+4b^2+c^2}\le\frac{1}{6b};\frac{1}{a^2+b^2+4c^2}\le\frac{1}{6c}\cdot\)
\(\Rightarrow M\le\frac{1}{6}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{ab+bc+ca}{abc}\cdot\frac{1}{6}\) \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow3\ge3\sqrt[3]{abc}\Rightarrow abc\le1\)
Theo BĐT \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
Khi đó \(M\le\frac{3}{1}\cdot\frac{1}{6}=\frac{1}{2}\)
Dấu "=" xảy ra tại \(a=b=c=1\)
P/S:Is that true ??
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=0\)
\(\Rightarrow\hept{\begin{cases}ab=-bc-ca\\bc=-ca-ab\\ca=-ab-bc\end{cases}}\)
Thay vào ta được: \(\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2+bc-ca-ab}=\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)
Tương tự: \(\frac{b^2}{b^2+2ca}=\frac{b^2}{\left(b-a\right)\left(b-c\right)}\) ; \(\frac{c^2}{c^2+2ab}=\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Rightarrow P=-\left[\frac{a^2}{\left(a-b\right)\left(c-a\right)}+\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(c-a\right)\left(b-c\right)}\right]\)
\(=-\left[\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right]\)
\(=\frac{\left(b-c\right)\left(a^2+bc-ca-ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(b-c\right)\left(a-b\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow ab+ac+bc=0\)
\(\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2-ab-ac+bc}=\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)
Tương tự: \(\frac{b^2}{b^2+2ac}=\frac{b^2}{\left(b-a\right)\left(b-c\right)};\frac{c^2}{c^2+2ac}=\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
\(VT\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2019}\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{x^2+z^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\) \(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\)
\(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\)
\(2\sqrt{2}VT\ge\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\)
\(2\sqrt{2}VT\ge\dfrac{4\left(x+y+z\right)^2}{2x+2y+2z}-\left(x+y+z\right)=x+y+z=\sqrt{2019}\)
\(\Rightarrow VT\ge\dfrac{\sqrt{2019}}{2\sqrt{2}}=\sqrt{\dfrac{2019}{8}}\) (đpcm)
