Ta có: \(\left\{{}\begin{matrix}\sqrt{a}+\sqrt{b}+\sqrt{c}=2\\\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}=\dfrac{1}{\sqrt{abc}}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)=4\\\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1\end{matrix}\right.\)
\(\Rightarrow a+b+c=2\)
Ta cần chứng minh:
\(b+c>4abc\)
\(\Leftrightarrow b+c-4\left(2-b-c\right)bc>0\)
\(\Leftrightarrow\left(b-4bc+4bc^2\right)+\left(c-4bc+4cb^2\right)>0\)
\(\Leftrightarrow\left(\sqrt{b}-2c\sqrt{b}\right)^2+\left(\sqrt{c}-2b\sqrt{c}\right)^2>0\) (đúng vì dấu = không xảy ra).
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