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I have below objective function: $$ \min_{x\geq 0} (\max \{ x, 1/x \} )$$

where, $x \geq 0$.

Can anyone please tell me how to proceed for this? Matlab says minimum value for $x$ is $0.1$, since I set boundary to .1, but thats not intuitive.

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    minimum value for x? I'm pretty sure you want minimum value for your function, which would be 1. X is free. – 2017-01-27
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    Could you please explain a little bit? I tried by plotting the two functions but I failed to understand the question basically. – 2017-01-27

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You have two functions, x and 1/x. If you plot them, you'll see that until x=1 the second function dominates (its plot is higher than the plot of the first function), and after that point, the first dominates.

So, max just chooses the dominating function.

The min here is where those functions meet - since 1/x decreases until x=1 and x increases after x=1.

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    Thanks a lot for clear and detailed explanation. – 2017-01-27
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Your function is 1/x from 0 to 1 and x from 1 to infinity.

Since 1/x monotonically decreases, consider if the minimum was at x<1. Then there would be a number a little bit less than it at an x a little bit greater than it.

Vice versa for x>1.

Now consider boundary conditions (namely x=0 and x=1).

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    Thank u. I kind of trying to convince myself. The answer seems to be x =1. – 2017-01-27