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.

  • 1
    $\begingroup$ @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 $x\to -\infty$. $\endgroup$
    – Irddo
    Jul 30, 2019 at 13:54

1 Answer 1


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 $f(x_1,y_1)=0$.

  • $\begingroup$ thanks for your help. I will look this right now! $\endgroup$
    – Irddo
    Jul 30, 2019 at 14:09

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