The least value of function $f(x) = ax + \frac{b}{x}($ where $,a > 0,b > 0,x > 0)$ is……………

The least value of function $f(x) = ax + \frac{b}{x}$ (where, $a > 0,b > 0,x > 0$) is $2\sqrt {ab}$.

${f^\prime }(x) = a - \frac{b}{{{x^2}}}$ and ${f^\prime }(x) = 0$

$\Rightarrow$ $a = \frac{b}{{{x^2}}}$

$\Rightarrow$ ${x^2} = \frac{b}{a} \Rightarrow x = \pm \sqrt {\frac{b}{a}}$

Now, ${f^{\prime \prime }}(x) = - b \cdot \frac{{( - 2)}}{{{x^3}}} = + \frac{{2b}}{{{x^3}}}$

At $x = \sqrt {\frac{b}{a}} ,$ ${f^{\prime \prime }}(x) = + \frac{{2b}}{{{{\left( {\frac{b}{a}} \right)}^{3/2}}}} = \frac{{ + 2b \cdot {a^{3/2}}}}{{{b^{3/2}}}}$

$= + 2{b^{ - 1/2}} \cdot {a^{3/2}} = + 2\sqrt {\frac{{{a^3}}}{b}} > 0$

Therefore the least value of $f(x),$ $f\left( {\sqrt {\frac{b}{a}} } \right) = a \cdot \sqrt {\frac{b}{a}} + \frac{b}{{\sqrt {\frac{b}{a}} }}$

$= a \cdot {a^{ - 1/2}} \cdot {b^{1/2}} + b \cdot {b^{ - 1/2}} \cdot {a^{1/2}}$

$= \sqrt {ab} + \sqrt {ab} = 2\sqrt {ab}$