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Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Tính \(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}\)
P= abc(\(\frac{1}{^{a^3}}\)+\(\frac{1}{b^3}\)+\(\frac{1}{c^3}\)) = abc[(\(\frac{1}{a}\)+\(\frac{1}{b}\))3+\(\frac{1}{c^3}\)-\(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)]=abc[(\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\))(....)- \(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)]
=abc.(- \(\frac{3}{a^2b}\)-\(\frac{3}{ab^2}\)) =-3(\(\frac{c}{a}\)+\(\frac{c}{b}\)) = -3c(\(\frac{1}{a}\)+\(\frac{1}{b}\)) = -3c.\(\frac{-1}{c}\)=3
P = 3
Đầu tiên,bạn cần chứng minh x + y + z = 0 thì x3 + y3 + z3 = 3xyz ( Bạn ko biết c/m thì hỏi nhé)
Thay\(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}=\frac{3}{abc}\)
\(\Rightarrow M=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}=abc\left(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}\right)=abc.\frac{3}{abc}=3\)
\(P=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(=\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\left(1\right)\)
Áp dụng BĐT AM-GM ta có: :
\(\frac{a}{a^2+b^2+c^2}+9a\left(a^2+b^2+c^2\right)\ge2\sqrt{9a^2}=6a\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{b}{a^2+b^2+c^2}+9b\left(a^2+b^2+c^2\right)\ge6b;\frac{c}{a^2+b^2+c^2}+9c\left(a^2+b^2+c^2\right)\ge6c\)
\(\Rightarrow\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}+9\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge6\left(a+b+c\right)\)
Theo BĐT Cauchy-Schwarz thì:
\(9\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge9\cdot\frac{\left(a+b+c\right)^2}{3}\cdot\left(a+b+c\right)=3\)
\(\Rightarrow\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}\ge6-3=3\)
Và \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{9}{ab+bc+ca}\ge\frac{9}{\frac{\left(a+b+c\right)^2}{3}}=27\)
Khi đó nhìn vào \(\left(1\right)\) thấy \(P\ge27+3=30\)
Xảy ra khi \(a=b=c=\frac{1}{3}\)
Áp dụng : x + y + z = 0 suy ra x3 + y3 + z3 = 3xyz
1/a + 1/2b + 1/3c = 0 = >... rồi biến đổi nhé
\(\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c^3}\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3.\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{3}{abc}\)
\(P=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)
trước hết ta chứng minh: nếu x+y+z=0 thì x3+y3+z3=3xyz
thật vậy, vì x+y+z=0 => z=-(x+y)
=> z3=-[x3+y3+3xy(x+y)]
=> x3+y3+z3=-3xy(x+y)=-3xy(-z)
=> x3+y3+z3=3xyz
áp dụng vào bài đã cho, ta suy ra: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{abc}\)
do đó \(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}=\frac{abc}{c^3}+\frac{bca}{a^3}+\frac{cab}{b^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)
vậy P=3
Chắc chắn là \(a^2+b^2+c^2=3\) rồi, thử \(a=b=c=\frac{1}{\sqrt{3}}\) là rõ
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ac}\ge\frac{\left(1+1+1\right)^2}{3+ab+bc+ca}\)
Ta có BĐT cơ bản \(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\frac{\left(1+1+1\right)^2}{3+ab+bc+ca}\ge\frac{\left(1+1+1\right)^2}{3+a^2+b^2+c^2}\)
\(\Rightarrow VT\ge\frac{\left(1+1+1\right)^2}{3+a^2+b^2+c^2}=\frac{9}{6}=\frac{3}{2}=VP\)
Đẳng thức xảy ra khi \(a=b=c=1\)
\(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3} +\frac{abc}{b^3}\)
\(=abc.\left(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}\right)\)Mà nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
thì \(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}=\frac{3}{abc}\)\(\Rightarrow P=abc.\frac{3}{abc}=3\)
Ta có :
\(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(=\frac{abc.3}{\left(abc\right)}=3\)