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a)\(VT=\sum_{cyc}\frac{ab^3+ab^2c+a^2bc}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\frac{\sum_{cyc}\left(ab^3+ab^2c+a^2bc\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}\)\(\le\frac{\sum_{cyc}ab\left(a^2+b^2\right)+abc\left(a+b+c\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}{\left(ab+bc+ca\right)^2}=\frac{a^2+b^2+c^2}{ab+bc+ca}=VP\)
\(VT=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{c^2b+abc+a^2b}+\frac{c^2}{a^2c+abc+b^2c}\)
Áp dụng bđt Cauchy dạng phân thức
\(\Rightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}\)
\(=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\frac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c\)
Chúc bạn học tốt !!!
Mình nghĩ đề nên cho a,b,c dương nếu không thì từ từ mình suy nghĩ
Đặt \(P=\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\)
Ta có:\(\frac{a-bc}{a+bc}=\frac{a-bc}{a\left(a+b+c\right)+bc}=\frac{a-bc}{\left(a+b\right)\left(a+c\right)}=\frac{\left(a-bc\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{\left(a-bc\right)\left(1-a\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\)
\(=\frac{a-a^2-bc+abc}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}=\frac{a-a^2-bc+abc}{1-a-b-c+ab+bc+ca-abc}=\frac{a-a^2-bc+abc}{ab+bc+ca-abc}\)
\(\Rightarrow P=\frac{a+b+c-a^2-b^2-c^2-ab-bc-ca+3abc}{ab+bc+ca-abc}\)
\(P=\frac{1-\left(a+b+c\right)^2+ab+bc+ca+3abc}{ab+bc+ca-abc}\)
\(P=\frac{ab+bc+ca+3abc}{ab+bc+ca-abc}=1+\frac{4abc}{ab+bc+ca-abc}\)
Cần cm:\(\frac{4abc}{ab+bc+ca-abc}\le\frac{1}{2}\)
\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)\ge9abc\)(đúng theo AM-GM)
"="<=>a=b=c=1/3
\(A=\frac{a^2}{a+bc}+\frac{b^2}{b+ca}+\frac{c^2}{c+ab}=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
\(=\frac{a^3}{a^2+ab+bc+ca}+\frac{b^3}{b^2+ab+bc+ca}+\frac{c^3}{c^2+ab+bc+ca}\)
\(=\frac{a^3}{\left(a+b\right)\left(c+a\right)}+\frac{b^3}{\left(b+c\right)\left(a+b\right)}+\frac{c^3}{\left(c+a\right)\left(b+c\right)}\)
đến đây áp dụng cô si 3 số là đc
Ta có \(\frac{a.1-bc}{a.1+bc}==\frac{a^2+ac}{a^2+ab+bc+ca}=\frac{a}{a+b}\)
Từ đó \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)
\(=-\left(\frac{a}{c-1}+\frac{b}{a-1}+\frac{c}{b-1}\right)=-\left(\frac{a^2}{ca-a}+\frac{b^2}{ab-b}+\frac{c^2}{bc-c}\right)\)
\(\le-\frac{\left(a+b+c\right)^2}{ab+bc+ca-\left(a+b+c\right)}=-\frac{1}{ab+bc+ca-1}\le-\frac{1}{\frac{\left(a+b+c\right)^2}{3}-1}=\frac{3}{2}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}.\)
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\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}=\frac{1}{\frac{1}{a}+\frac{1}{b}}=\frac{1}{\frac{1}{b}+\frac{1}{c}}=\frac{1}{\frac{1}{c}+\frac{1}{a}}\)
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Leftrightarrow a=b=c\)
=>M = 1