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If $x,y$ are positive reals satisfying $x^2+y^2=1$, then the minimum value of $x+y+\frac{1}{xy}$ is?

Attempt 1

I used the Lagrange multipliers method which ended up being cumbersome.

Attempt 2

I applied the AM-GM inequality to obtain $\frac{1}{xy}\ge2$ and $x+y\le\sqrt2$ after which I'm lost again.

Im looking for hints/better approaches to the problem.

DatBoi
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  • @Saad It does. I was pretty darn sure that this question would be present on this site, but I never found it when I was writing the post! – DatBoi Jul 12 '21 at 05:55
  • (In addition to that: since the expression is symmetric then $x=y$ gives either a local minimum or a local maximum. Since $x\to 0$ or $y\to0$ give maxima, it's 'sensible' that $x=y$ is a minimum; you should be able to prove it straightforwardly. Showing that the finction is increasing from $x=y$ to $x=0$ shouldn't be hard either.) – Steven Stadnicki Jul 12 '21 at 05:56
  • AM-GM the way you are applying will not work as $x, y$ can be negative – Math Lover Jul 12 '21 at 06:00
  • It has been mentioned in the question that they are positive reals @MathLover(though the linked question doesnt) – DatBoi Jul 12 '21 at 06:01
  • @StevenStadnicki noted – DatBoi Jul 12 '21 at 06:02
  • OK I see that now. Can you share how you approached using Lagrange Multiplier method? It should have given you point $x = y$ and minimum of $2 + \sqrt2$ which is also evident by symmetry. – Math Lover Jul 12 '21 at 06:14
  • @MathLover I did the usual things and ended up with an equation with squares and square roots(as a result of substitution of $y$) – DatBoi Jul 12 '21 at 06:15
  • You surely get $x = y$ as one of the points when you apply Lagrange Multiplier method. – Math Lover Jul 12 '21 at 06:18
  • Let us continue this discussion in chat. – DatBoi Jul 12 '21 at 06:19
  • @Saad the question that you have tagged this as a duplicate of does not state that $a, b$ are positive real. Even the accepted solution does not call out, which would be wrong without $a, b \gt 0$ constraint. There are comments asking OP if they had forgotten $a, b \gt 0$ constraint which OP never cares to respond to or edit the question. – Math Lover Jul 12 '21 at 06:31
  • How did lagrange multipliers turn out cumbersome???? – tryst with freedom Jul 12 '21 at 07:22
  • @Buraian check out the discussion in the chat – DatBoi Jul 12 '21 at 07:30
  • I think I got the problem, the issue is that you have to use some algebra tricks to keep the calculations in Lagrange simple. Direct application is of lagrange can be dificult at times – tryst with freedom Jul 12 '21 at 07:50

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