Calculus/real analysis problem from a Differential geometry exercice

I found a calculus/real analysis problem from a differential geometry exercice of surfaces in $$\mathbb{R}^3$$.

Consider $$U=\mathbb{R}\times(0,1/2)$$ and let be $$f:U\subset\mathbb{R}^2\to \mathbb{R}$$ given by $$f(x,y)=(2x-\sqrt{4x^2+1})y\sqrt{1-4y^2}+\sqrt{x^2+y^2}.$$

I ploted the graph of $$(x,y,f(x,y))$$ in $$\mathbb{R}^3$$ and I saw that $$f\geq 0$$, but I don't realized how prove that $$f\geq 0$$ on $$U$$ by hand.

First, I worked with the inequality $$(2x-\sqrt{4x^2+1})y\sqrt{1-4y^2}+\sqrt{x^2+y^2}\geq 0$$ and the assumption that $$y\in(0,1/2)$$ but nothing. Secondly, I tried a change of variables to polar coordinates, but nothing too.

Some help for this? I will appreciate, thanks.

• @Matematleta yes! I think that you are wrong about the limit. For $y=1/4$ we get that $f(x,1/4)=1/8\, \left( 2\,x-\sqrt {4\,{x}^{2}+1} \right) \sqrt {3}+1/4\,\sqrt { 16\,{x}^{2}+1}$. For this, $f(x,1/4)\to +\infty$ when $x\to -\infty$. Jul 30, 2019 at 13:54

Here's a brute-force proof . . .

For $$(x,y)\in U$$, \begin{align*} &f(x,y)\ge 0\\[4pt] \iff\;&(2x-\sqrt{4x^2+1})y\sqrt{1-4y^2}+\sqrt{x^2+y^2}\ge 0\\[4pt] \iff\;&\sqrt{x^2+y^2}\ge \bigl(\!\sqrt{4x^2+1}-2x\bigr)y\sqrt{1-4y^2}\\[4pt] \iff\;&x^2+y^2\ge \bigl(\!\sqrt{4x^2+1}-2x\bigr)^2y^2(1-4y^2)\\[4pt] \iff\;&\frac{x^2+y^2}{y^2(1-4y^2)}\ge \bigl(\!\sqrt{4x^2+1}-2x\bigr)^2\\[4pt] \iff\;&\frac{x^2+y^2}{y^2(1-4y^2)}\ge 8x^2-4x\sqrt{4x^2+1}+1\\[4pt] \iff\;&\frac{x^2+y^2}{y^2(1-4y^2)}-(8x^2+1)\ge -4x\sqrt{4x^2+1}\\[4pt] \iff\;&\frac{x^2-8x^2y^2+32x^2y^4+4y^4}{y^2(1-4y^2)}\ge -4x\sqrt{4x^2+1}\\[4pt] \iff\;&\frac{x^2(1-8y^2+16y^4)+(16x^2y^4+4y^4)}{y^2(1-4y^2)}\ge -4x\sqrt{4x^2+1}\\[4pt] \iff\;&\frac{x^2(1-4y^2)^2+(16x^2y^4+4y^4)}{y^2(1-4y^2)}\ge -4x\sqrt{4x^2+1}\\[4pt] \end{align*} which is true (with strict inequality) if $$x\ge 0$$ since the $$\text{LHS}$$ is positive, and the $$\text{RHS}$$ is nonpositive.

Thus, it remains to consider the case $$x < 0$$.

For $$x < 0$$, we have \begin{align*} &\frac{x^2(1-4y^2)^2+(16x^2y^4+4y^4)}{y^2(1-4y^2)}\ge -4x\sqrt{4x^2+1}\\[4pt] \iff\;&\left(\frac{x^2(1-4y^2)^2+(16x^2y^4+4y^4)}{y^2(1-4y^2)}\right)^2\ge \left(-4x\sqrt{4x^2+1}\right)^2\\[4pt] \iff\;&\left(\frac{x^2(1-4y^2)^2+(16x^2y^4+4y^4)}{y^2(1-4y^2)}\right)^2\ge 16x^2(4x^2+1)\\[4pt] \iff\;&\left(\frac{x^2(1-4y^2)^2+(16x^2y^4+4y^4)}{y^2(1-4y^2)}\right)^2-16x^2(4x^2+1)\ge 0\\[4pt] \end{align*} which is true since the $$\text{LHS}$$ factors as $$\left(\frac{x^2-8x^2y^2-4y^4}{y^2(1-4y^2)}\right)^2$$

Therefore $$f(x,y)\ge 0$$, for all $$(x,y)\in U$$.

Moreover, we have \begin{align*} &f(x,y)=0\\[4pt] \iff\;&x < 0\;\;\;\text{and}\;\;\;x^2-8x^2y^2-4y^4=0\\[4pt] \iff\;&x < 0\;\;\;\text{and}\;\;\;x^2=\frac{4y^4}{1-8y^2}\\[4pt] \iff\;&1-8y^2 > 0\;\;\;\text{and}\;\;\;x=-\frac{2y^2}{\sqrt{1-8y^2}}\\[4pt] \iff\;&0 < y < \frac{\sqrt{2}}{4}\;\;\;\text{and}\;\;\;x=-\frac{2y^2}{\sqrt{1-8y^2}}\\[4pt] \end{align*} For example, if $$y_1={\large{\frac{1}{4}}}$$ and $$x_1=-{\large{\frac{1}{8}}}\sqrt{2}$$, then $$f(x_1,y_1)=0$$.

• thanks for your help. I will look this right now! Jul 30, 2019 at 14:09