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Ta có \(a+b+c=abc\Leftrightarrow\dfrac{a+b+c}{abc}=1\) \(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
Lại có \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\Leftrightarrow2^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) (đpcm)
\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)
Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)
Cộng vế với vế:
\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
Ta có: \(a=b+c\Rightarrow c=a-b\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{\dfrac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}}=\sqrt{\dfrac{b^2\left(a-b\right)^2+a^2\left(a-b\right)^2+a^2b^2}{a^2b^2c^2}}=\sqrt{\dfrac{b^4+a^2b^2-2ab^3+a^4+a^2b^2-2a^3b+a^2b^2}{a^2b^2c^2}}=\sqrt{\dfrac{\left(a^2+b^2\right)^2-2ab\left(a^2+b^2\right)+a^2b^2}{a^2b^2c^2}}=\sqrt{\dfrac{\left(a^2+b^2-ab\right)^2}{a^2b^2c^2}}=\left|\dfrac{a^2+b^2-ab}{abc}\right|\)
=> Là một số hữu tỉ do a,b,c là số hữu tỉ
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
\(\Rightarrow\dfrac{1}{a}=\left(\dfrac{1}{2}-\dfrac{1}{b}\right)+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{b-2}{2b}+\dfrac{c-2}{2c}\)
Dễ dàng chứng minh \(\dfrac{b-2}{2b},\dfrac{c-2}{2c}\) là các số dương.
Áp dụng BĐT Cauchy cho 2 số dương ta có:
\(\dfrac{b-2}{2b}+\dfrac{c-2}{2c}\ge2\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{4bc}}=\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{bc}}\)
\(\Rightarrow\dfrac{1}{a}\ge\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{bc}}\left(1\right)\)
CMTT ta có: \(\left\{{}\begin{matrix}\dfrac{1}{b}\ge\sqrt{\dfrac{\left(c-2\right)\left(a-2\right)}{ca}}\left(2\right)\\\dfrac{1}{c}\ge\sqrt{\dfrac{\left(a-2\right)\left(b-2\right)}{ab}}\left(3\right)\end{matrix}\right.\)
\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow\dfrac{1}{abc}\ge\dfrac{\left(a-2\right)\left(b-2\right)\left(c-2\right)}{abc}\)
\(\Rightarrow\left(a-2\right)\left(b-2\right)\left(c-2\right)\le1\left(đpcm\right)\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=b=c\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{matrix}\right.\Leftrightarrow a=b=c=3\)
Đồng thời chỉ ra phương pháp nhé!!