Ta thấy: \(5x+2\sqrt{x}+22=\left(5x+2\sqrt{x}+\frac{1}{5}\right)+\frac{109}{5}=\left[\left(\sqrt{5x}\right)^2+2.\sqrt{5x}.\frac{\sqrt{5}}{5}+\left(\frac{\sqrt{5}}{5}\right)^2\right]+\frac{109}{5}=\left(\sqrt{5x}+\frac{\sqrt{5}}{5}\right)^2+\frac{109}{5}>0\text{ với mọi x}\ge0,xe-4\)
Với x\(\ge0\), \(xe-4\) thì 2(x+4)\(\ge2\left(0+4\right)=8>0\)
Do đó:\(\frac{5x+2\sqrt{x}+22}{2\left(x+4\right)}>0\)