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1) ta có: a(b^2 -1)(c^2 -1)+b(a^2 -1)(c^2 -1)+c(a^2-1)(b^2-1)
=(ab^2 -a)(c^2-1)+(ba^2 -b)(c^2-1)+(ca^2-c)(b^2-1)
đén đây nhân bung ra hết rồi rút gọn và thay a+b+c=abc là đc
1. Ta có : \(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)
Tương tự : \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\); \(\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\). Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=9\)
\(9\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=49\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=49\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=49\)
\(P=\dfrac{1}{abc}+\dfrac{1}{a^2+b^2+c^2}=\dfrac{a+b+c}{abc}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\left(1\right)\)
\(\)\(\left\{{}\begin{matrix}a+b+c=1\\\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{9}{ab+bc+ac}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow P\ge\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{1}{2\left(ab+bc+ac\right)}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{17}{2\left(ab+bc+ac\right)}\)
\(\Rightarrow P\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{17}{2\left(ab+bc+ac\right)}\)
\(\Rightarrow P\ge9+\dfrac{17}{2\left(ab+bc+ac\right)}\)
mà \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\)
\(\Rightarrow P\ge9+\dfrac{17}{2.\dfrac{1}{3}}=9+\dfrac{17.3}{2}=\dfrac{18+17.3}{2}=\dfrac{69}{2}\)
\(\Rightarrow Min\left(P\right)=\dfrac{69}{2}\)
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+c^3=\dfrac{3}{abc}\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{3}{abc}=0\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2-\dfrac{1}{c}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{c^2}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=0\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{1}{ab}-\dfrac{1}{bc}-\dfrac{1}{ca}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b=c\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\)
Đề bài thiếu, cần thêm dữ liệu "a;b;c phân biệt"
Khi đó \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\)
\(\Rightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
Ta có : \(a^2+b^2\ge2ab\Rightarrow a^2+b^2-ab\ge ab\)
\(\Rightarrow\dfrac{1}{a^2-ab+b^2}\le\dfrac{1}{ab}=\dfrac{abc}{ab}=c\) ( do $abc=1$ )
Tương tự ta có :
\(\dfrac{1}{b^2-bc+c^2}\le a\)
\(\dfrac{1}{c^2-ab+a^2}\le b\)
Cộng vế với vế các BĐT trên có :
\(\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{b^2-bc+c^2}+\dfrac{1}{c^2-ac+a^2}\le a+b+c\)
Dấu "=" xảy ra khi $a=b=c$
\(VT=\dfrac{1}{a^2+b^2-ab}+\dfrac{1}{b^2+c^2-bc}+\dfrac{1}{c^2+a^2-ca}\)
\(VT\le\dfrac{1}{2ab-ab}+\dfrac{1}{2bc-bc}+\dfrac{1}{2ca-ca}=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{a+b+c}{abc}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Đặt \(\frac{a}{b^2}=x,\frac{b}{c^2}=y,\frac{c}{a^2}=z\).
\(\Rightarrow xyz=\frac{abc}{a^2b^2c^2}=\frac{1}{abc}=1\)
Theo bài ra ta có : \(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\)
\(\Leftrightarrow x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow x+y+z=xy+yz+xz\)
\(\Leftrightarrow\left(xy-x-y+1\right)-1+z\left(x+y-1\right)=0\)
\(\Leftrightarrow\left(xy-x-y+1\right)+z\left(x+y-1-xy\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)-z\left(x-1\right)\left(y-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(1-z\right)=0\)
\(\Leftrightarrow\frac{a-b^2}{b^2}.\frac{b-c^2}{c^2}.\frac{a^2-c}{a^2}=0\)
\(\Leftrightarrow\left(a-b^2\right)\left(b-c^2\right)\left(c-a^2\right)=0\)
Ta có đpcm
Vi a + b + c = 1 nên bt tương đương với \(P=abc\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
Ta có : \(P=abc\left(a+b+c\right)\left(a^2+b^2+c^2\right)\le\frac{1}{3}\left(ab+bc+ca\right)^2\left(a^2+b^2+c^2\right)\)( 1 )
Mặt khác :\(\left(ab+bc+ca\right)^2\left(a^2+b^2+c^2\right)\le\left(\frac{\left(a+b+c\right)^2}{3}\right)^3=\frac{1}{27}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow P\le\frac{1}{3}.\frac{1}{27}=\frac{1}{81}\)
Dấu "=" xảy ra <=> a = b = c = 1/3
Vậy maxP = 1/81 <=> a = b = c = 1/3