Question

Answer 1

Applying Cauchy-Schwarz ; we have : 

\(A=\dfrac{\sqrt{x-3}}{2x}=\dfrac{\sqrt{\left(x-3\right)3}}{2x\sqrt{3}}\le\dfrac{x}{4x\sqrt{3}}=\dfrac{1}{4\sqrt{3}}\) 

\(B=\dfrac{\sqrt{2x-5}}{9x}=\dfrac{\sqrt{\left(2x-5\right)5}}{9x\sqrt{5}}\le\dfrac{x}{9x\sqrt{5}}=\dfrac{1}{9\sqrt{5}}\)

\(C=\dfrac{x+8}{\sqrt{x-1}}=\dfrac{x-1+9}{\sqrt{x-1}}=\sqrt{x-1}+\dfrac{9}{\sqrt{x-1}}\ge6\)

\(D=\dfrac{2x+6}{\sqrt{2x-1}}=\sqrt{2x-1}+\dfrac{7}{\sqrt{2x-1}}\ge2\sqrt{7}\)