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\(P=\frac{ab+c}{\left(a+b\right)^2}.\frac{bc+a}{\left(b+c\right)^2}.\frac{ac+b}{\left(c+a\right)^2}\)
\(P=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ac+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(P=\frac{ab+ac+bc+c^2}{\left(a+b\right)^2}.\frac{ab+bc+ac+a^2}{\left(b+c\right)^2}.\frac{ab+bc+ac+b^2}{\left(a+c\right)^2}\)
\(P=\frac{\left(a+c\right)\left(b+c\right)}{\left(a+b\right)^2}.\frac{\left(a+b\right)\left(a+c\right)}{\left(b+c\right)^2}.\frac{\left(a+b\right)\left(b+c\right)}{\left(a+c\right)^2}=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2}=1\)
\(a+b+c=1\Rightarrow\hept{\begin{cases}ab+c=ab+c\left(a+b+c\right)\\bc+a=bc+a\left(a+b+c\right)\\ca+b=ca+b\left(a+b+c\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=ab+ca+bc+c^2\\bc+a=bc+a^2+ab+ac\\ca+b=ca+ab+b^2+bc\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=\left(b+c\right)\left(a+c\right)\\bc+a=\left(a+c\right)\left(a+b\right)\\ca+b=\left(b+c\right)\left(a+b\right)\end{cases}}\)
\(\Rightarrow P=\frac{\left(b+c\right)\left(a+c\right)}{\left(a+b\right)^2}.\frac{\left(a+c\right)\left(a+b\right)}{\left(b+c\right)^2}.\frac{\left(b+c\right)\left(a+b\right)}{\left(c+a\right)^2}=1\)
\(P=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ca+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(=\frac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)
a) Ta có : \(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự : \(b^2+1=\left(b+a\right)\left(b+c\right)\) ; \(c^2+1=\left(c+a\right)\left(c+b\right)\)
Suy ra \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
Vậy \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)
b) Ta có ; \(a^2+2bc-1=a^2+2bc-\left(ab+bc+ac\right)=a^2-ab+bc-ac=a\left(a-b\right)-c\left(a-b\right)\)
\(=\left(a-b\right)\left(a-c\right)\)
Tương tự : \(b^2+2ac-1=\left(a-b\right)\left(c-b\right)\) ; \(c^2+2ab-1=\left(a-c\right)\left(b-c\right)\)
Suy ra \(\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ab-1\right)=\left(a-b\right)^2.\left(c-a\right)^2.\left[-\left(b-c\right)^2\right]\)
Vậy : \(B=\frac{-\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)}=-1\)
Dễ dàng chứng minh được:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)với \(x,y>0\)(1)
Dấu bằng xảy ra \(\Leftrightarrow x=y>0\)
Ta có:
\(\frac{a}{bc\left(a+1\right)}=\frac{a}{abc+bc}=\frac{a}{ab+bc+ca+bc}=\frac{a}{\left(ab+bc\right)+\left(bc+ca\right)}\)
Áp dụng (1), ta được:
\(\frac{1}{ab+bc}+\frac{1}{bc+ca}\ge\frac{4}{\left(ab+bc\right)+\left(bc+ca\right)}\)
\(\Leftrightarrow\frac{1}{4\left(ab+bc\right)}+\frac{1}{4\left(bc+ca\right)}\ge\frac{1}{ab+bc+bc+ca}\)
\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{ab+bc+bc+ca}\)
\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{bc\left(a+1\right)}\left(2\right)\)
Dấu bằng xảy ra \(\Leftrightarrow b=c>0\)
Chúng minh tương tự, ta được:
\(\frac{b}{4}\left(\frac{1}{ab+ca}+\frac{1}{bc+ca}\right)\ge\frac{b}{ca\left(b+1\right)}\left(3\right)\)
Dấu bằng xảu ra \(\Leftrightarrow a=c>0\).
\(\frac{c}{4}\left(\frac{1}{ac+ab}+\frac{1}{ab+bc}\right)\ge\frac{c}{ab\left(c+1\right)}\left(4\right)\)
Từ (2), (3) và (4), ta được:
\(\frac{a}{bc\left(a+1\right)}+\frac{b}{ca\left(b+1\right)}+\frac{c}{ab\left(c+1\right)}\le\)\(\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ac}\right)+\frac{b}{4}\left(\frac{1}{ac+bc}+\frac{1}{ac+ab}\right)\)\(+\frac{c}{4}\left(\frac{1}{ab+bc}+\frac{1}{ab+ac}\right)\)
\(\Leftrightarrow P\le\frac{1}{4}.\left(\frac{a}{ab+bc}+\frac{c}{ab+bc}\right)+\frac{1}{4}\left(\frac{a}{bc+ac}+\frac{b}{bc+ac}\right)\)\(+\frac{1}{4}\left(\frac{b}{ab+ac}+\frac{c}{ab+ac}\right)\)
\(\Leftrightarrow P\le\frac{a+c}{4\left(ab+bc\right)}+\frac{a+b}{4\left(bc+ac\right)}+\frac{b+c}{4\left(ab+ac\right)}\)
\(\Leftrightarrow P\le\frac{a+c}{4b\left(a+c\right)}+\frac{a+b}{4c\left(a+b\right)}+\frac{b+c}{4a\left(b+c\right)}\)
\(\Leftrightarrow P\le\frac{1}{4b}+\frac{1}{4c}+\frac{1}{4a}\)
\(\Leftrightarrow P\le\frac{1}{4}\left(\frac{ab+bc+ca}{abc}\right)\)
\(\Leftrightarrow P\le\frac{1}{4}.\frac{abc}{abc}=\frac{1}{4}.1=\frac{1}{4}\)( vì \(ab+bc+ca=abc\))
Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\ab+bc+ca=abc\end{cases}}\Leftrightarrow a=b=c=3\)
Vậy \(minP=\frac{1}{4}\Leftrightarrow a=b=c=3\)
\(P=\frac{ab+c.1}{\left(a+b\right)^2}.\frac{bc+a.1}{\left(b+c\right)^2}.\frac{ca+b.1}{\left(c+a\right)^2}\)
\(P=\frac{ab+c\left(a+b+c\right)}{\left(a+b\right)^2}.\frac{bc+a\left(a+b+c\right)}{\left(b+c\right)^2}.\frac{ca+b\left(a+b+c\right)}{\left(c+a\right)^2}\)
\(P=\frac{ab+ca+bc+c^2}{\left(a+b\right)^2}.\frac{bc+a^2+ab+ac}{\left(b+c\right)^2}.\frac{ca+ab+b^2+bc}{\left(c+a\right)^2}\)
\(P=\frac{\left(a+c\right)\left(b+c\right)}{\left(a+b\right)^2}.\frac{\left(a+c\right)\left(a+b\right)}{\left(b+c\right)^2}.\frac{\left(a+b\right)\left(b+c\right)}{\left(c+a\right)^2}=1\)