Find the minimum of $f:(0,\infty)^2\rightarrow\mathbb{R},(x,y)\mapsto 1/x+1/y+2x+2y$.

Question

Find the minimum of $f:(0,\infty)^2\rightarrow\mathbb{R},(x,y)\mapsto 1/x+1/y+2x+2y$ making use of the following theorem:

If $f:A\rightarrow\mathbb{R}$ is convex and $\xi$ has a directional derivative in every direction with $\nabla f(\xi)=0$, then $f(\xi)=\min\{f(x):x\in A\}$.

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Now if I understand correctly I need to show that 1. $f$ is convex, 2. that $f$ has a directional derivative in every direction, 3. find $(x,y)$ such that $\nabla f(x,y)=0$. I am having trouble with step 1 in particular and a question for step 2. I will first show step 2 and 3 since those are the easiest to get out of the way.

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First, regarding step 2: Is it correct that if I show $\nabla f(x,y)$ exists, that this implies that a directional derivative in every direction exists? Making use of partial differentiation I get

$\nabla f(x,y)=(2-\frac{1}{x^2}, 2-\frac{1}{y^2})$.

For step 3 I must find $(x,y)$ such that the vector above equals zero, which is $x=\sqrt{1/2}$ and $y=\sqrt{1/2}$.

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For step 1 I tried to show making use of the definition. I do not believe this to be sufficient. If I simplify $$f(tx_1+(1-t)x_2),ty_1+(1-t)y_2)-tf(x_1,x_2)-(1-t)f(x_2,y_2)$$ I get

$$\frac{1}{t(x_1-x_2)+x_2} + \frac{1}{t(y_1-y_2+y_2} + t(1/x_1-1/y_1+1/x_2+1/y_2) - 1/x_2 -1/y_2.$$

How do I argue that this is $\geq 0$?

Answer 1

Solution by a different method: $min \{f(x,y):x,y>0\} =min \{\frac 1 x +2x, x>0\}+min \{\frac 1 y +2y, y>0\}=2min \{\frac 1 x +2x, x>0\}$.

The derivative of $\frac 1 x +2x$ is $2-\frac 1 {x^{2}}$ which vanishes at $x=1/\sqrt 2$ and the minimum value is $2\sqrt 2$. hence the answer is $4\sqrt 2$.

Answer 2

Assuming $x > 0, y > 0$ we have

$$ 1/x+1/y+2x+2y \ge 4\left(\frac 1x\frac 1y (2x)(2y)\right)^{\frac 14} = 4(4)^{\frac 14} = 4\sqrt 2 $$

Answer 3

Just to answer OP's question in original setup, instead of proposing a new solution altogether (since OP's goal seems to be learning about directional derivatives, and not minimizing a particular function):

1. To show that $f$ is strictly convex, it is simplest in this case to prove that the Hessian matrix is positive definite (which it is)
2. To prove that $f$ has directional derivative in every direction, you could show first that the partial derivatives in every point of the domain are continuous in that point (which they are). This implies $f$ is differentiable in that point, which in turn implies that the directional derivatives exist in all directions.
3. For step 3, what you did is correct