\(A=\dfrac{1}{\dfrac{16}{a^2}}+\dfrac{1}{\dfrac{4}{b^2}}+\dfrac{1}{c^2}\)
\(A\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{a^2+b^2+c^2}\)
\(A\ge\dfrac{49}{\dfrac{16}{1}}=\dfrac{49}{16}\)
"="<=>\(c^2=2b^2=4a^2\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(A=\left(\dfrac{1}{16a^2}+\dfrac{1}{4b^2}+\dfrac{1}{c^2}\right)\left(a^2+b^2+c^2\right)\)
\(\ge\left(\sqrt{\dfrac{1}{16a^2}\cdot a^2}+\sqrt{\dfrac{1}{4b^2}\cdot b^2}+\sqrt{\dfrac{1}{c^2}\cdot c^2}\right)^2\)
\(=\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2=\dfrac{49}{16}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{\dfrac{1}{16a^2}}{a^2}=\dfrac{\dfrac{1}{4b^2}}{b^2}=\dfrac{\dfrac{1}{c^2}}{c^2}\\a^2+b^2+c^2=1\end{matrix}\right.\)\(\Leftrightarrow a=\dfrac{1}{\sqrt{7}};b=\sqrt{\dfrac{2}{7}};c=\dfrac{2}{\sqrt{7}}\)
