#### 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 )