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\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c};c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{c^3+b^3+d^3}\left(1\right)\\ \text{Đặt }\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;b=ck;c=dk\\ \Rightarrow a=bk=ck^2=dk^3\\ \Rightarrow\dfrac{a}{d}=k^3\\ \text{Mà }\dfrac{a}{b}=k\Rightarrow\dfrac{a^3}{b^3}=k^3\\ \Rightarrow\dfrac{a}{d}=\dfrac{a^3}{b^3}\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)
Ta có:
\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+c\left(a+b\right)\right)-abc\)
\(=\left(a+b\right)ab+\left(a+b\right)^2c+abc+c^2\left(a+b\right)-abc\)
\(=\left(a+b\right)\left(ab+c^2+c\left(a+b\right)\right)\)
\(=\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Đồng thời:
\(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự:
\(b^2+1=\left(a+b\right)\left(b+c\right)\)
\(c^2+1=\left(a+c\right)\left(b+c\right)\)
Từ đó:
\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}=1\)
Vì: \(0\le a\le b\le c\le1\) nên:
\(\left(a-1\right).\left(b-1\right)\ge0\Leftrightarrow ab-a-b+1\ge0\Leftrightarrow ab+1\ge a+b\)
\(\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\) (1)
\(\left(a-1\right).\left(c-1\right)\ge0\Leftrightarrow ac-a-c+1\ge0\Leftrightarrow ac+1\ge a+c\)
\(\Leftrightarrow\dfrac{1}{ac+1}\le\dfrac{1}{a+c}\Leftrightarrow\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\) (2)
\(\left(b-1\right).\left(c-1\right)\ge0\Leftrightarrow bc-b-c+1\ge0\Leftrightarrow bc+1\ge b+c\)
\(\Leftrightarrow\dfrac{1}{bc+1}\le\dfrac{1}{b+c}\Leftrightarrow\dfrac{a}{bc+1}\le\dfrac{a}{b+c}\) (3)
Cộng vế với vế của (1)(2) và (3) ta được:
\(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{2a+2b+2c}{a+b+c}\)
\(\Leftrightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{2.\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ac+1}\le2\left(đpcm\right)\)
\(\left\{{}\begin{matrix}b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\\c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}\left(1\right)\)
Và \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)
Áp dụng t/c dtsbn:
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\Rightarrow a=b=c\)
\(\Rightarrow P=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a\cdot a\cdot a}=\dfrac{8a^3}{a^3}=8\)
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+a+c-b+b+c-a}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)
\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)
b^2=ac
=>b/a=c/b=k
=>b=ak; c=bk=ak*k=ak^2
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+a^2k^2}{a^2k^2+a^2k^4}=\dfrac{1}{k^2}\)
\(\dfrac{a}{c}=\dfrac{a}{ak^2}=\dfrac{1}{k^2}\)
=>\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)