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Lời giải:
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ac=0\)
\(\Rightarrow 0=(ab+bc+ac)^2=a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=0\)
Hiển nhiên \(a^2b^2,b^2c^2,c^2a^2\geq 0\rightarrow a^2b^2+b^2c^2+c^2a^2\geq 0\)
Dấu bằng xảy ra khi \(ab=bc=ac=0\)
Vì vậy, không thể có TH \(a,b,c\neq 0\), do đó đề bài sai.
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Áp dụng BĐT Cauchy - schwarz dưới dạng engel ta có :
\(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\ge\dfrac{9}{2\left(a+b+c\right)}=\dfrac{4,5}{a+b+c}>\dfrac{3}{a+b+c}\)
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\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
Cộng vế: \(VT+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow VT\ge\dfrac{a+b+c}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
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Từ đkđb
\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Hớ hớ bài này mình cũng làm rồi.
Ta có: (a+b+c)2=a2+b2+c2
<=> a2+b2+c2+2(ab+bc+ca)=a2+b2+c2
<=>2(ab+bc+ca)=0
<=>ab+bc+ca=0
\(\Leftrightarrow\dfrac{ab+bc+ca}{abc}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=>\(\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\)
=> \(\dfrac{1}{a^3}+\dfrac{3}{a^2b}+\dfrac{3}{ab^2}+\dfrac{1}{b^3}=-\dfrac{1}{c^3}\)
=>\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{3}{ab}.\left(-\dfrac{1}{c}\right)=\dfrac{3}{abc}\)
=> Đpcm.
Xét \(a+b=0\) thì ta có ĐPCM
Xét \(b=c\)
\(\Rightarrow a=2c\)
Ta chứng minh:
\(\dfrac{8c^3+c^3}{8c^3+c}=\dfrac{2c+c}{2c+c}\)
\(\Leftrightarrow1=1\) đúng
Xét \(\left\{{}\begin{matrix}a+b\ne0\\b\ne c\end{matrix}\right.\)
Ta chứng minh:
\(\dfrac{a^3+b^3}{a^3+c^3}=\dfrac{a+b}{a+c}\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+c\right)\left(a^2-ac+c^2\right)}=\dfrac{a+b}{a+c}\)
\(\Leftrightarrow\dfrac{a^2-ab+b^2}{a^2-ac+c^2}=1\)
\(\Leftrightarrow a^2-ab+b^2=a^2-ac+c^2\)
\(\Leftrightarrow a\left(c-b\right)=\left(c-b\right)\left(c+b\right)\)
\(\Leftrightarrow a=c+b\) đúng
Vậy ta có ĐPCM