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Đặt x = \(\dfrac{1}{a}\); y = \(\dfrac{1}{b}\); z = \(\dfrac{1}{c}\); x + y + z = 0 (vì \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\))
x = -(y + z)
x3 + y3 + z3 - 3xyz = - (y + z)3 + y3 + z3 - 3xyz
-(y3 + 3y2z + 3y2z2 + z3) + y3 + z3 - 3xyz = -3yz(y + z + x) = -3yz . 0 = 0
Từ x3 + y3 +z3 - 3xyz = 0 \(\Leftrightarrow\) x3 + y3 +z3 = 3xyz
Do đó P = \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{3}{abc}=3\)
Nếu \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) thì \(\)P = \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}=3\)
Ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+3.\dfrac{1}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+3.\dfrac{1}{ab}.\left(-\dfrac{1}{c}\right)=0\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}-\dfrac{3}{abc}=0\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\Leftrightarrow abc.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{3abc}{abc}\Leftrightarrow\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=3\Leftrightarrow P=3\)Vậy khi \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) thì P=3
\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)
Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bđt Cauchy ta có :
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=3\)
Theo C.B.S thì
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{9}{ab+bc+ac}\)
\(\Rightarrow\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ac}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}\)
Lại theo CBS thì
\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}=9\)mà \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{7}{ab+bc+ac}\ge21\)
\(\Rightarrow\)\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}\)\(\)\(\ge21+9=30\)
vậy Min = 30 khi a = b = c = 1/3
Bạn xem lời giải tại đây:
https://hoc24.vn/cau-hoi/cho-abcge0a2b2c21cmr-dfracc1abdfracb1acdfraca1bcge1.1019784090594
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
theo de bai ta co \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\) suy ra ab+bc+ac=abc
\(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ac}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)
nên vt =\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(a+c\right)\left(c+b\right)}\)
nx \(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\) >= \(\dfrac{3a}{4}\)
ttu vt>= \(\dfrac{3\left(a+b+c\right)}{4}-\left(\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{a+b}{8}+\dfrac{b+c}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\right)\) =\(\dfrac{a+b+c}{4}\)
dau = say ra a=b=c=3
Lời giải:
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow \frac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow ab+bc+ac=0\)
\(\Rightarrow ab+bc=-ac\). Khi đó:
\(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}=\frac{(ab)^3+(bc)^3+(ca)^3}{a^2b^2c^2}\)
\(=\frac{(ab+bc)^3-3(ab)^2bc-3ab(bc)^2+(ca)^3}{a^2b^2c^2}\)
\(=\frac{(-ac)^3-3ab^2c(ab+bc)+(ca)^3}{a^2b^2c^2}\)
\(=\frac{-3ab^2c(ab+bc)}{a^2b^2c^2}=\frac{-3ab^2c.(-ac)}{a^2b^2c^2}=3\)
@Akai Haruma Góp ý nhỏ ạ. Chị nên đọc đề trước khi trả lời, Bản chất bài này đề sai,Cái đầu là ab chứ ko phải ac. Em nghĩ chị nên có từ sửa đề ở đầu bài giải