(Non-symmetric) inequality

Let us consider the function $mxy +\dfrac{1}{xy}$ , where $m>0$ . Then, since $x+y=1\implies y=1-x$ ,

the function can be reduced to one in a single variable $x$ :

$mx(1-x)+\dfrac{1}{x(1-x)}=\dfrac{m}{4}-m(x-\frac{1}{2})^2+\dfrac{4}{1-4(x-\frac{1}{2})^2}$

$=\dfrac{m+16}{4}+\dfrac{(x-\frac{1}{2})^2\left (4m(x-\frac{1}{2})^2+16-m\right )}{1-4(x-\frac{1}{2})^2}$ .

Now $x\leq 1\implies 4(x-\frac{1}{2})^2\leq 1\implies 1-4(x-\frac{1}{2})^2\geq 0$ .

Hence, for $0<m\leq 16$ , $mxy+\dfrac{1}{xy}\geq \dfrac{m+16}{4}$ .

In this case, $0<m=5<16$ , and the minimum is $\dfrac{5+16}{4}=\dfrac{21}{4}=\boxed {5.25}$ .

I will use differentiation to get the answer.

Let $A=5xy+\frac{1}{xy} = 5x(1-x) + \frac{1}{x(1-x)} = 5x(1-x) + \frac{1}{x} +\frac{1}{1-x}$ . Then

$\frac{dA}{dx} = 5-10x -\frac{1}{x^2} + \frac{1}{\left(1-x\right)^2} = -5\left(2x-1\right) + \frac{2x-1}{x^2 \left(x-1\right)^2} = \left(2x-1\right) \left( \frac{1}{x^2 \left(x-1\right)^2}-5 \right)$

As $0<x<1$ , $\frac{dA}{dx}=0$ if and only if $x=\frac{1}{2}$ .

Now $\frac{d^2A}{dx^2}=-10 + \frac{2}{x^3} + \frac{2}{\left(1-x\right)^3} >0 \, \, \forall \, 0<x<1$

Thus $A$ is minimum if $x=\frac{1}{2}$ . Hence the minimum value of $A$ is $5\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\frac{1}{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)} = \boxed{5.25}$ .

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*Remarks:

Interestingly, if $0<m<16$ and the positive real numbers $x$ and $y$ are such that $x+y=1$ , then the minimum value of $mxy+\frac{1}{xy}$ always occurs at $x=y=\frac{1}{2}$ .

$Using\ AM-GM\\ \frac{x+y}{2} \geq \sqrt[]{xy}\\ (\frac{1}{2})^2 \geq xy\ \ \ \mathbf{(1)}\\ \frac{5}{4} \geq 5xy\ \ \ \mathbf{(2)}\\ using\ (1)\ to\ -1\\ (\frac{1}{4})^{-1} \geq xy^{-1}\\ 4 \geq \frac{1}{xy}\ \ \ \mathbf{(3)}\\ Sum\ (2)\ and\ (3)\\ \frac{21}{4} \geq 5xy + \frac{1}{xy} \\ min\left ( 5xy + \frac{1}{xy} \right ) = \frac{21}{4}$