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Ta có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Leftrightarrow ab+bc+ca=0\)
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Áp dụng :
\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc\)
\(\Rightarrow a^6+b^6+c^6\)
\(=\left(a^2+b^2+c^2\right)\left(a^4+b^4+c^2-a^2b^{^2}-b^2c^2-c^2a^2\right)+3a^2b^2c^2\)
\(\Leftrightarrow a^6+b^6+c^6=3a^2b^2c^2\)
Ta CM BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}},a+b\ge2\sqrt{ab}\)( co si với a,b>0)
Suy ra \(\left(\frac{1}{a}+\frac{1}{b}\right)\left(a+b\right)\ge4\RightarrowĐPCM\)\(\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)
a/Áp dụng (1) có
\(\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\left(2\right)\).Tương tự ta cũng có:
\(\frac{1}{b+c+2a}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\left(3\right),\frac{1}{c+a+2b}\le\frac{1}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\left(4\right)\)
Cộng (2),(3) và (4) có \(VT\le\frac{1}{4}.\left(6+6\right)=3\left(ĐPCM\right)\)
b/Áp dụng (1) có:
\(\frac{1}{3a+3b+2c}=\frac{1}{\left(a+b+2c\right)+2\left(a+b\right)}\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{2\left(a+b\right)}\right)\left(5\right)\)
Tương tự có: \(\frac{1}{3a+2b+3c}\le\frac{1}{4}\left(\frac{1}{a+c+2b}+\frac{1}{2\left(a+c\right)}\right)\left(6\right)\)
\(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{2a+b+c}+\frac{1}{2\left(b+c\right)}\right)\left(7\right)\)
Cộng (5),(6) và (7) có:
\(VT\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{a+c+2b}+\frac{1}{2a+b+c}+\frac{1}{2}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\right)\le\frac{1}{4}.9=\frac{3}{2}\)
\(P=\frac{2a+3b+3c-1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2c+1}{2017+c}\)
\(=\frac{6047-a}{2015+a}+\frac{6048-b}{2016+b}+\frac{6049-c}{2017+c}\)
\(=\frac{8062}{2015+a}+\frac{8064}{2016+b}+\frac{8066}{2017+c}-3\)
\(\ge\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{2015+2016+2017+a+b+c}-3=\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{8064}-3\)
Dấu = xảy ra khi ....
Ta có : \(P=\frac{2a+3b+3c+1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2c-1}{2017+c}\)
\(\Rightarrow P+3=\frac{2a+3b+3c+1}{2015+a}+1+\frac{3a+2b+3c}{2016+b}+1+\frac{3a+3b+2c-1}{2017+c}+1\)
\(=\frac{3a+3b+3c+2016}{2015+a}+\frac{3a+3b+3c+2016}{2016+b}+\frac{3a+3b+3c+2016}{2017+c}\)
\(=\left(3a+3b+3c+2016\right)\left(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\right)\)
\(=4.2016\left(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\right)\) \(\left(a+b+c=2016\right)\)
\(=8064.\left(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\right)\)
Vì a ; b ; c dương , áp dụng BĐT phụ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\), ta có :
\(\frac{1}{2015+a}+\frac{1}{2016+b}+\frac{1}{2017+c}\ge\frac{9}{2015+2016+2017+a+b+c}=\frac{9}{8064}\)
\(\Rightarrow P+3\ge8064.\frac{9}{8064}=9\) \(\Rightarrow P\ge6\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2015+a=2016+b=2017+c\\a+b+c=2016\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1=c+2\\a+b+c=2016\end{matrix}\right.\)
\(\Leftrightarrow a=673;b=672;c=671\)
Vậy ...
\(\frac{1}{3a}+\frac{1}{2b}+\frac{1}{c}=\frac{1}{3a+2b+c}\)
\(\Leftrightarrow\frac{1}{3a}+\frac{1}{2b}=\frac{1}{3a+2b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{1}{3a}+\frac{1}{2b}=\frac{c-\left(3a+2b+c\right)}{\left(3a+2b+c\right)c}\)
\(\Leftrightarrow\frac{3a+2b}{6ab}=\frac{-\left(3a+2b\right)}{3ac+2bc+c^2}\)
\(\Leftrightarrow\left(3a+2b\right)\left(3ac+2bc+c^2\right)+\left(3a+2b\right)6ab=0\)
\(\Leftrightarrow\left(3a+2b\right)\left(3ac+2bc+c^2+6ab\right)=0\)
\(\Rightarrow\left(3a+2b\right)\left(2b+c\right)\left(c+3a\right)=0\) (đpcm)