Áp dụng BĐT Cauchy - Schwarz ta có :
\(VT=\frac{1}{\sqrt{a}}+\frac{3}{\sqrt{b}}+\frac{8}{\sqrt{3c+2a}}\)
\(=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{b}}+\frac{8}{\sqrt{3c+2a}}\)
\(\ge\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{2\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}\)
\(=\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}+\frac{\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}\)
\(\ge\frac{\left(1+2+1+2+2\right)^2}{2\sqrt{3c+2a}+3\sqrt{b}+\sqrt{a}}\)
\(\ge\frac{64}{\sqrt{\left(1+2^2+3\right)\left(a+2a+3c+3b\right)}}\)
\(=\frac{64}{\sqrt{24\left(a+c+b\right)}}=\frac{16\sqrt{2}}{\sqrt{3\left(a+b+c\right)}}=VF\)
Chúc bạn học tốt !!!
Mình nghĩ là:
a = 1
b = 2
c = 4
