Do \(a,b,c>\dfrac{25}{4}\Rightarrow\) các mẫu số đều dương
Áp dụng BĐT Cauchy:
\(M\ge3\sqrt[3]{\dfrac{abc}{\left(2\sqrt{b}-5\right)\left(2\sqrt{c}-5\right)\left(2\sqrt{a}-5\right)}}\)
\(\Rightarrow M\ge3\sqrt[3]{\dfrac{5^3.abc}{5\left(2\sqrt{b}-5\right).5\left(2\sqrt{c}-5\right).5\left(2\sqrt{a}-5\right)}}\)
Ta có: \(\left\{{}\begin{matrix}5\left(2\sqrt{a}-5\right)\le\dfrac{\left(5+2\sqrt{a}-5\right)^2}{4}=a\\5\left(2\sqrt{b}-5\right)\le\dfrac{\left(5+2\sqrt{b}-5\right)^2}{4}=b\\5\left(2\sqrt{c}-5\right)\le\dfrac{\left(5+2\sqrt{c}-5\right)^2}{4}=c\end{matrix}\right.\)
\(\Rightarrow M\ge3\sqrt[3]{\dfrac{5^3.abc}{abc}}=3.5=15\)
\(\Rightarrow M_{min}=15\) khi \(a=b=c=25\)
