Áp dụng bđt Svác xơ, ta có:
\(A\ge\dfrac{\left(\sqrt{2x}+\sqrt{3y}+\sqrt{4z}\right)^2}{2\left(4x^2+9y^2+16z^2\right)}\)\(=\dfrac{2x+3y+4z+2\left(\sqrt{6xy}+\sqrt{12yz}+\sqrt{8xz}\right)}{2}\)\(\ge\dfrac{1+2\left(3\sqrt[3]{\sqrt{576x^2y^2z^2}}\right)}{2}\)(BĐT Cô-si)\(\ge\dfrac{1+6}{2}=\dfrac{7}{2}\)
Vậy Amin=\(\dfrac{7}{2}\Leftrightarrow\)\(\left\{{}\begin{matrix}\dfrac{2x}{9y^2+16z^2}=\dfrac{3y}{4x^2+16z^2}=\dfrac{4z}{4x^2+9y^2}\\\sqrt{6xy}=\sqrt{12yz}=\sqrt{8xz}\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{3}{2}y=2z\)
Viết lại bài toán: Cho \(a^2+b^2+c^2=1\). Tìm max \(\sum\dfrac{a}{b^2+c^2}\)
với a=2x, b=3y, c=4z.
Áp dụng BĐT AM-GM:
\(a\left(b^2+c^2\right)=\dfrac{1}{\sqrt{2}}\sqrt{2a^2\left(1-a^2\right)\left(1-a^2\right)}\le\dfrac{1}{\sqrt{2}}\sqrt{\dfrac{8}{27}}=\dfrac{2}{3\sqrt{3}}\)
Do đó \(VT\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)
Vậy \(A_{Min}=\dfrac{3\sqrt{3}}{2}\)
