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Ta xét hiệu :
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ac+a^2}\right)\)
\(=a-b+b-c+c-a=0\)
Do đó : \(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ac+a^2}=1006\)
Khi đó \(M=2\cdot1006=2012\)
\(\frac{bc+a^2}{a+b}+\frac{ac+b^2}{b+c}+\frac{ab+c^2}{a+c}\ge\)a+b+c
<=>\(\frac{bc+a^2}{a+b}-a+\frac{ac+b^2}{b+c}-b+\frac{ab+c^2}{a+c}-c\ge0\)
<=>\(\frac{b\left(c-a\right)}{a+b}+\frac{c\left(a-b\right)}{b+c}+\frac{a\left(b-c\right)}{a+c}\ge0\)
<=>\(\frac{b\left(b+c\right)\left(a+c\right)\left(a-c\right)}{\left(a+b\right)\left(c+c\right)\left(a+c\right)}\)+\(\frac{c\left(a+c\right)\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a\left(a+b\right)\left(b-c\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
<=>\(\frac{b^2c^2-b^2a^2+bc^3-a^2bc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a^3c-ab^2c+c^2a^2-b^2c^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a^2b^2-a^2c^2+ab^3-abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
<=>\(\frac{bc^3+a^3c+ab^3-a^2bc-ab^2c-abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
<=>\(\frac{2bc^3+2a^3c+2ab^3-2a^2bc-2ab^2c-2abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)>=0
<=>\(\frac{bc\left(c-a\right)^2+ac\left(a-b\right)^2+ab\left(b-c\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(đung voi moi a,b,c >0)
Dấu ''='' xay ra khi a=b=c
a, \(M=\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ca+2c+2}\)
\(=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}\)
\(=\frac{1+b+bc}{1+b+bc}=1\)
\(\Rightarrow M=1\)
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)
\(\Rightarrow\frac{bcx+acy+abz}{abc}=0\)
\(\Rightarrow bcx+acy+abz=0\)
Lại có:\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\frac{bcx+acy+abz}{xyz}=4\)(bình phương hai vế)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4\)(Vì \(bcx+acy+abz=0\))
Từ (1) \(\Rightarrow bcx+acy+abz=0\)
Gọi \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\left(2\right)\)
Từ (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=4-\left(\frac{abz+acy+bcx}{xyz}\right)\)
\(=4\)
\(b,\frac{ab}{a^2+b^2+c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
Từ \(a+b+c=0\Rightarrow a+b=-c\Rightarrow a^2+b^2-c^2=-2ab\)
Tương tự \(b^2+c^2-a^2=-2bc\)và \(c^2+a^2-b^2=-2ac\)
\(\Rightarrow\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=\frac{1}{-2}+\frac{1}{-2}+\frac{1}{-2}\)
\(=-\frac{3}{2}\)
1) Theo bđt AM-GM,ta có: \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}=a\)
Suy ra \(\frac{a^2}{b+c}\ge a-\frac{b+c}{4}\)
Thiết lập hai BĐT còn lại tương tự và cộng theo vế ta có đpcm
Áp dụng bất đẳng thức Bunyakovsky, ta được: \(\Sigma_{cyc}\frac{ab}{a^2+bc+ca}=\Sigma_{cyc}\frac{ab\left(b^2+bc+ca\right)}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\Sigma_{cyc}\frac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)
Ta có: \(\Sigma_{cyc}\frac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}=\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}=\frac{ab^3+bc^3+ca^3+2.a\sqrt{ab}.c\sqrt{ab}+2.a\sqrt{bc}.b\sqrt{bc}+2.c\sqrt{ca}.b\sqrt{ca}}{\left(ab+bc+ca\right)^2}\le\frac{ab^3+bc^3+ca^3+a^3b+abc^2+a^2bc+b^3c+c^3a+ab^2c}{\left(ab+bc+ca\right)^2}=\frac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{\left(ab+bc+ca\right)^2}=\frac{a^2+b^2+c^2}{ab+bc+ca}\)
Đẳng thức xảy ra khi a = b = c
Áp dụng BĐT Bunhiacopxki:
\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)
\(\Rightarrow\frac{ab}{a^2+bc+ca}\le\frac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)
Tương tự: \(\frac{bc}{b^2+ca+ab}\le\frac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\) ; \(\frac{ac}{c^2+ab+bc}\le\frac{ac\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
Cộng vế với vế:
\(VT\le\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}\)
\(VT\le\frac{ab^3+bc^3+ca^3+2.a\sqrt{ab}.c\sqrt{ab}+2a\sqrt{bc}.b\sqrt{bc}+2c\sqrt{ac}.b\sqrt{ac}}{\left(ab+bc+ca\right)^2}\)
\(VT\le\frac{ab^3+bc^3+ca^3+a^3b+abc^2+b^3c+a^2bc+ac^3+ab^2c}{\left(ab+bc+ca\right)}=\frac{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}{\left(ab+bc+ca\right)^2}\)
\(VT\le\frac{a^2+b^2+c^2}{ab+bc+ca}\)
Dấu "=" xảy ra khi \(a=b=c\)
ta có: a + b + c = 0 => a + b = - c => (a+b)2 = (-c)2 => a2 + 2ab + b2 = c2 => a2 + b2 - c2 = -2ab
chứng minh tương tự, ta có: b2 + c2 -a2 = -2bc; c2 + a2 - b2 = -2ac
\(B=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
\(B=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ac}\)
\(B=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)
vậy B là số hữu tỉ