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Sửa đề: \(\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2021}\\abc=2021\end{cases}}\) thì \(M=\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)\) là số chính phương
Ta có: \(\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2021}\\abc=2021\end{cases}}\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{abc}\Rightarrow ab+bc+ca=1\left(abc\ne0\right)\)
Khi đó ta có: \(\hept{\begin{cases}1+a^2=ab+bc+ca+a^2=\left(a+b\right)\left(a+c\right)\\1+b^2=\left(b+c\right)\left(b+a\right)\\1+c^2=\left(c+a\right)\left(c+b\right)\end{cases}}\)
Nhân vế với vế ta được:
\(M=\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
=> M là số chính phương
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow ab+bc+ca=1\)
\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)\)
\(\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(b+c\right)\left(c+a\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
Áp dụng bất đẳng thức Cô-si ta có :
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge3\sqrt[3]{\frac{1}{a^3b^3c^3}}=\frac{3}{abc}\)
Dấu = xảy ra khi \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\) Hay \(a=b=c\) ( đề cho )
Vậy ta có đpcm : \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
a) Ta có: \(\frac{a^2}{a+b}-\frac{b^2}{a+b}+\frac{b^2}{b+c}-\frac{c^2}{b+c}+\frac{c^2}{c+a}-\frac{a^2}{c+a}\) \(=a-b+b-c+c-a=0\)
\(\Rightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}=\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}\)
\(\Rightarrow2\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\right)=\frac{a^2}{a+b}+\frac{b^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{b+c}+\frac{c^2}{c+a}+\frac{a^2}{c+a}\)\(\ge\frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ca}{c+a}\)
\(\Rightarrowđpcm\)
Dấu "=" \(\Leftrightarrow a=b=c\)
b) \(a^2b^2\left(a^2+b^2\right)=\frac{1}{2}\cdot ab\cdot2ab\cdot\left(a^2+b^2\right)\le\frac{1}{2}\cdot\frac{\left(a+b\right)^2}{4}\cdot\frac{\left(2ab+a^2+b^2\right)^2}{4}=2\)
Dấu "=" \(\Leftrightarrow a=b=1\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2 \Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
<=> \(ab+bc+ac=0\Leftrightarrow\frac{ab+ac+bc}{abc}=0\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
<=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\frac{1}{c^3}\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+3.\frac{1}{a^2}.\frac{1}{b}+3.\frac{1}{a}.\frac{1}{b^2}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}\left(\frac{-1}{c}\right)=0\Leftrightarrow\)dpcm
Từng ý nhé !!!
\(P=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(\frac{1}{abc}.3abc=3\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
Xét \(a+b+c=0\) ta có :\(\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
\(Q=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\frac{b^2}{\left(b+c\right)\left(b-c\right)-a^2}+\frac{c^2}{\left(c+a\right)\left(c-a\right)-b^2}\)
\(=\frac{a^2}{-ac+bc-c^2}+\frac{b^2}{-ab+ac-a^2}+\frac{c^2}{-bc+ab-b^2}\)
\(=\frac{a^2}{-c\left(a+c\right)+bc}+\frac{b^2}{-a\left(a+b\right)+ac}+\frac{c^2}{-b\left(c+b\right)+ab}\)
\(=\frac{a^2}{bc+bc}+\frac{b^2}{ac+ac}+\frac{c^2}{ab+ab}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{1}{2abc}\left(a^3+b^3+c^3\right)=\frac{1}{2abc}.3abc=\frac{3}{2}\)
Xét \(a=b=c\) ta có :
\(Q=\frac{a^2}{a^2-a^2-a^2}+\frac{b^2}{b^2-b^2-b^2}+\frac{c^2}{c^2-c^2-c^2}=-1-1-1=-3\)
\(a\left(a^2-bc\right)+b\left(b^2-ca\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow a^3-abc+b^3-abc+c^3-abc=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
Mà \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}\Rightarrow}a=b=c\)
Vậy \(P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
@Nguyễn Việt Lâm
Đề bài sai mà, từ giả thiết suy ra \(\left[{}\begin{matrix}a=c\\ac=b^2\end{matrix}\right.\)
Trường hợp đầu tiên, chọn \(a=c=2;b=1\Rightarrow a^2+b^2+c^2=9\)
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