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\(\dfrac{ab}{6+a-c}+\dfrac{bc}{6+b-a}+\dfrac{ca}{6+c-b}=\dfrac{ab}{2a+b}+\dfrac{bc}{2b+c}+\dfrac{ca}{2c+a}\)
Mà ta có:
\(\dfrac{2a+b}{ab}=\dfrac{2}{b}+\dfrac{1}{a}=\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{a}\ge\dfrac{9}{2b+a}\)
\(\Rightarrow\dfrac{ab}{2a+b}\le\dfrac{2b+a}{9}\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{bc}{2b+c}\le\dfrac{2c+b}{9}\\\dfrac{ca}{2c+a}\le\dfrac{2a+c}{9}\end{matrix}\right.\)
Cộng 3 cái trên vế theo vế ta được
\(\dfrac{ab}{2a+b}+\dfrac{bc}{2b+c}+\dfrac{ca}{2c+a}\le\dfrac{3\left(a+b+c\right)}{9}=\dfrac{3.6}{9}=2\)
\(\left(a+b+c\right)=\dfrac{1}{2}\Leftrightarrow\left(a+b+c\right)^2=\dfrac{1}{4}\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=\dfrac{1}{4}\)
Ta có: \(ab+bc+ac=\left(a^2+b^2+c^2+2ab+2bc+2ac\right)-\left(a^2+b^2+c^2+ab+bc+ac\right)=\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{1}{12}\)
\(a^2+b^2+c^2=\dfrac{1}{6}-\left(ab+bc+ac\right)=\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)
Suy ra: \(a^2+b^2+c^2=ab+bc+ac\Leftrightarrow a=b=c\)
\(P=\dfrac{3}{2}\)
p/s làm lih tih k chắc đâu:v
\(M=\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+bc+c^2}+\frac{c^3+a^3}{c^2+ac+a^2}\)
\(=\left(\frac{a^3+b^3}{a^2+ab+b^2}-b+a\right)+\left(\frac{b^3+c^3}{b^2+bc+c^2}-c+b\right)+\left(\frac{c^3+a^3}{c^2+ac+a^2}-a+c\right)\)
\(=2\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}\right)\)
\(=2....\) ( đề thiếu )
\(\dfrac{a^3-b^3}{ab^2}+\dfrac{b^3-c^3}{bc^2}+\dfrac{c^3-a^3}{ca^2}\ge0\)
\(\Leftrightarrow\dfrac{a^2}{b^2}-\dfrac{b}{a}+\dfrac{b^2}{c^2}-\dfrac{c}{b}+\dfrac{c^2}{a^2}-\dfrac{a}{c}\ge0\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge\dfrac{2a}{c}\\\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{2b}{a}\\\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge\dfrac{2c}{b}\end{matrix}\right.\)
Cộng 3 cái vế theo vế rồi rút gọn cho 2 ta được ĐPCM
a/d vào công thức a^3+b^3+b^3=3abc( khi a+b+c=0)
ta đc 1/a+1/b+1/c=0
=> (1/a)^3+(1/b)^3+(1/c)^3=3. (1/abc)
lại có S=\(\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)
=abc (\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\))
=3.\(\dfrac{abc}{abc}\)=1
chúc bạn học tốt ^ ^
Dễ CM : nếu x+y+z=0 thì x^3+y^3+z^3=3xyz
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
\(S=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\\ =abc.\dfrac{1}{abc}=1\)
Lời giải:
Ta có: \(B=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}\)
\(B=\frac{(bc)^3+(ca)^3+(ab)^3}{a^2b^2c^2}\)
Vì \(ab+bc+ac=0\Rightarrow bc+ac=-ab\). Do đó:
\((bc)^3+(ca)^3+(ab)^3=(bc+ca)^3-3bc.ca(bc+ca)+(ab)^3\)
\(=(-ab)^3-3bc.ca(-ab)+(ab)^3\)
\(=3bc.ca.ab=3a^2b^2c^2\)
Suy ra : \(B=\frac{3a^2b^2c^2}{a^2b^2c^2}=3\)