Question
Given m>0 , prove \dfrac{24}{m}+6m\geq24
Given m>0 , prove \dfrac{24}{m}+6m\geq24
According to Inequality of arithmetic and geometric means AM–GM inequality,
\dfrac{24}{m}+6m\geq 2\cdotp \sqrt{\dfrac{24}{m} \cdotp 6m } =24
Only if \dfrac{24}{m}=6m , two sides of inequality are equal.
Solve the equation, we get
m=2 (cancel -2 because m>0 )