$\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2\sqrt{2}}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{\frac{c}{2}}}+\frac{1}{\sqrt{\frac{c}{2}}}\geq \frac{(1+1+1+1)^2}{\sqrt{a}+\sqrt{b}+\sqrt{\frac{c}{2}}+\sqrt{\frac{c}{2}}}$ (Theo BĐT Cauchy Shwars dạng Engel).
Mặt khác ta có BĐT: $a+b+c+d\leq 2\sqrt{(a^2+b^2+c^2+d^2)}$.
Do đó: $\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2\sqrt{2}}{\sqrt{c}}\geq \frac{(1+1+1+1)^2}{\sqrt{a}+\sqrt{b}+\sqrt{\frac{c}{2}}+\sqrt{\frac{c}{2}}}\geq \frac{16}{2(a+b+\frac{c}{2}+\frac{c}{2})}=8(a+b+c+d)$.
Dấu "=" xảy ra $\Leftrightarrow a=b=\frac{c}{2}$.
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