Find a minimum of $f(x,y)$ without the isoperimetric inequality

Question

Let $H=\{(x,y)\in R^2: y>0\}$, and let $f:H\to R$ be defined by $$f(x,y)=\frac{2+\sqrt{(1+x)^2+y^2}+\sqrt{(1-x)^2+y^2}}{\sqrt y}.$$ Show that $f$ attains its minimum on $H$ at a unique point and find that point.

The numerator is the perimeter of the triangle with vertices $(\pm 1,0), (x,y)$. The denominator is the root of its area. I can use the isoperimetric inequality for triangles (and that the equality is achieved for regular triangles) to solve this. But could I solve this if I didn't know the inequality? Dealing with partial derivatives seems to be too cumbersome. Or is the purpose of this problem to prove the isoperimetric inequality? If so, how do I prove it (and that the equality is achieved for regular triangles)?

Answer 1

Hint:

Use that $$ a^2+b^2 \geq {1\over 2}(a+b)^2$$

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Then :

$$\sqrt{(1+x)^2+y^2} \geq {1\over \sqrt{2}}(1+x+y)$$ and

$$\sqrt{(1-x)^2+y^2} \geq {1\over \sqrt{2}}(1-x+y)$$ so

$$f(x,y)\geq \frac{2+ \sqrt{2}(1+y)}{\sqrt y}\geq {2\sqrt{\sqrt{2}(2+\sqrt{2})y}\over \sqrt{y}} =2\sqrt{2\sqrt{2}+2}$$ Last inequality is between AG and GM ($a+b\geq 2\sqrt{ab}$).

Answer 2

$f(-x,y)=f(x,y)$ so $x=0$ is an axis of symmetry.

From the form of $f(x,y)$ you can see that it will increase if $|x|$ increases, so the minimum if it exists will be reached on the line $x=0$.

$f(0,y)=\dfrac{2+2\sqrt{1+y^2}}{\sqrt{y}}$

This is not to complicated to derivate $\dfrac{y^2-1-\sqrt{1+y^2}}{y\sqrt{y}\sqrt{1+y^2}}$

Annulation of denominator arises for $(y^2-1)^2=1+y^2\iff y^4=3y^2\implies y=\sqrt{3}$ on the domain $H$.

$f(0,\sqrt{3})=2\times 3^{\frac 34}\approx 4.559...$

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@Angle

In your answer you get $2\sqrt{2\sqrt{2}+2}\approx 4.394...$ which is less what I obtained, so one of us did wrong.

I understood you were obtaining $y=1+\sqrt{2}$ but $f(0,1+\sqrt{2})\approx 4.650...$, I'm confused about the values of $(x,y)$ for the minimum ?