Question

a,b,c>0. CM: \(\dfrac{1}{\sqrt{a}}\) + \(\dfrac{3}{\sqrt{b}}\) + \(\dfrac{8}{\sqrt{3c+2a}}\) \(\ge\) \(\dfrac{16\sqrt{2}}{\sqrt{3\left(a+b+c\right)}}\)

Answer 1

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT=\dfrac{1}{\sqrt{a}}+\dfrac{3}{\sqrt{b}}+\dfrac{8}{\sqrt{3c+2a}}\)

\(=\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{2}{\sqrt{b}}+\dfrac{8}{\sqrt{3c+2a}}\)

\(\ge\dfrac{4}{\sqrt{a}+\sqrt{b}}+\dfrac{2\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}\)

\(=\dfrac{4}{\sqrt{a}+\sqrt{b}}+\dfrac{\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}+\dfrac{\left(1+2\right)^2}{\sqrt{3c+2a}+\sqrt{b}}\)

\(\ge\dfrac{\left(1+2+1+2+2\right)^2}{2\sqrt{3c+2a}+3\sqrt{b}+\sqrt{a}}\)

\(\ge\dfrac{64}{\sqrt{\left(1+2^2+3\right)\left(a+2a+3c+3b\right)}}\)

\(=\dfrac{64}{\sqrt{24\left(a+c+b\right)}}=\dfrac{16\sqrt{2}}{\sqrt{3\left(a+b+c\right)}}=VP\)