Prove that $f(x, y) \le 3 $ for $x \ge 0, y > 0$

Question

Let $x \ge 0, y>0$ and \begin{align*} f(x,y)&=\sqrt{\dfrac{y}{y+x^2}}+4\sqrt{\dfrac{y}{(y+(x+1)^2)(y+(x+3)^2)}}\\[6pt] &\qquad +4\sqrt{\dfrac{y}{(y+(x-1)^2)(y+(x-3)^2)}}. \end{align*} Prove that $f(x,y) \le 3$.

I can prove when $x\ge 2, f(x,y) < 3$, but $f(x,y)=3$ when $x=0,y=3$, so the key part is $x\le 2 $

I found when $x<2, f'_x(y \ge 1.5) <0$ but I can't prove it as it has high degree equations.

And for $x<2,y<1.5$, I have no idea how to prove $f(x,y)<3$

Answer 1

No clue without an overview. With help of restrictions as found already by the OP, a contour plot / isoline chart has been produced for the function at hand:

The $y$-axis is in $\color{green}{\mbox{green}}$. Our viewport is:

  xmin := -6 ; xmax := 6;
  ymin := 0 ; ymax := 12;

The are 27 contour levels nivo, defined by:

  for geval := 1 to 27 do
  begin
    nivo := geval/9.1;

The higher the level, the more black. The lower the level, the more white. The pixels in $\color{red}{red}$ are close to the conjectured maximum $=3$ . They are defined by:

const
  eps : double = 0.0002;
  if (3-f < eps) and (3-f > 0) then

It is seen that the red spot is inside an area with very black lines, meaning that the function $f(x,y)$ is increasing there, towards the red spot.

Analysis. There is a mirror symmetry in the $y$-axis, as is clear from the figure but also from some algebra. Thus for all real $x$ and for all real $y > 0$ we have: $$ f(x,y) = f(-x,y) $$ It follows that: $$ \frac{\partial f(x,y)}{\partial x} = - \frac{\partial f(-x,y)}{\partial x} \quad \Longrightarrow \quad \left.\frac{\partial f(x,y)}{\partial x}\right|_{x=0} = - \left.\frac{\partial f(-x,y)}{\partial x}\right|_{x=0} \\ \Longrightarrow \quad \left.\frac{\partial f(x,y)}{\partial x}\right|_{x=0} = 0 $$ Meaning that all function gradients near the $y$-axis are tangent to it. Since the gradient vectors are perpendicular to the isolines, this can be observed in the figure as well. Now a sufficient condition for having a stationary point (maximum or minimum eventually) is that the total gradient be zero. But we already have $\,\partial f / \partial x = 0\,$ at the $y$-axis. Therefore substitute $\,x=0\,$ in $\,f(x,y)\,$ and only consider function values at the $y$-axis: $$ f(0,y)=1+8\sqrt{\frac{y}{(y+1)(y+9)}} $$ Extremes are found for: $$ \frac{d f(0,y)}{dy} = 0 \quad \Longrightarrow \\ \frac{1}{2} \left[ {\frac {1}{ \left( y+1 \right) \left( y+9 \right) }}-{ \frac {y}{ \left( y+1 \right) ^{2} \left( y+9 \right) }}-{\frac {y}{ \left( y+1 \right) \left( y+9 \right) ^{2}}} \right] \sqrt{\frac{\left( y+1 \right) \left( y+9 \right)}{y}} = 0 \\ \Longrightarrow \quad (y+1)(y+9) - y(y+9) - y(y+1) = -y^2 +9 = 0 \quad \Longrightarrow \quad y = \pm 3 $$ With the restriction $\,y > 0$ . Hence the only extreme at $\,(y,f(0,y))$ is $(3,3)\,$ and it is a maximum.
Picking up the comment by nbubis, prove that $f(0,2) < 3$ and $f(0,4) < 3$: $$ f(0,2)=1+8\sqrt{\frac{2}{(2+1)(2+9)}}=1+\sqrt{\frac{128}{33}}<1+\sqrt{\frac{128}{32}}\\ f(0,4)=1+8\sqrt{\frac{4}{(4+1)(4+9)}}=1+\sqrt{\frac{256}{65}}<1+\sqrt{\frac{256}{64}} $$

Answer 2

Here is a proof:

Using Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} [f(x, y)]^2 &\le \left(\frac{3y}{4(x^2 + y)} + \frac{y}{y + (x + 1)^2} + \frac{y}{y + (x - 1)^2}\right)\\ &\qquad \times \left(\frac43 + \frac{16}{y + (x + 3)^2} + \frac{16}{y + (x - 3)^2}\right). \end{align*} It suffices to prove that \begin{align*} &9 \ge \left(\frac{3y}{4(x^2 + y)} + \frac{y}{y + (x + 1)^2} + \frac{y}{y + (x - 1)^2}\right)\\ &\qquad \times \left(\frac43 + \frac{16}{y + (x + 3)^2} + \frac{16}{y + (x - 3)^2}\right) \end{align*} or \begin{align*} &16\,{y}^{5}+ \left( 91\,{x}^{2}+64 \right) {y}^{4}+ \left( 204\,{x}^{4 }+60\,{x}^{2}-672 \right) {y}^{3}\\ &\qquad + \left( 226\,{x}^{6}-612\,{x}^{4}- 1038\,{x}^{2}+576 \right) {y}^{2}\\ &\qquad + \left( 124\,{x}^{8}-1148\,{x}^{6}+ 2388\,{x}^{4}-612\,{x}^{2}+1296 \right) y\\ &\qquad +27\,{x}^{10}-540\,{x}^{8}+ 3186\,{x}^{6}-4860\,{x}^{4}+2187\,{x}^{2} \ge 0. \tag{1} \end{align*}

My proof of (1) is not nice. nexu@AoPS gave a very nice SOS (Sum of Squares) proof of (1): \begin{align*} &16\,y^5+\left( 91\,x^2+64 \right) y^4+\left( 204\,x^4+60\,x^2-672 \right) y^3\\ &\qquad +\left( 226\,x^6-612\,x^4-1038\,x^2+576 \right) y^2 \\ &\qquad +\left( 124\,x^8-1148\,x^6+2388\,x^4-612\,x^2+1296 \right) y\\ &\qquad +27\,x^{10}-540\,x^8+3186\,x^6-4860\,x^4+2187\,x^2 \\ =\,& \frac{y\left( 38x^4+55x^2y-209x^2-6y^2-8y+78 \right) ^2}{38}\\ &\qquad +\frac{x^2\left( 27x^4+43x^2y-270x^2-20y^2-21y+243 \right) ^2}{27} \\ &\qquad +\frac{800039x^2y^2(y-3)^2}{28356}+\frac{1843x^2y(y-3)^2}{43}+175y^2(x^2+y-3)^2\\ &\qquad +\frac{142709197y(x^2+y-3)^2}{1310482} +\frac{x^2y^2(4726x^2+3583y-10749)^2}{255204}\\ &\qquad +\frac{y\left( 7683x^2y-23769x^2+857y^2-5862y+9873 \right) ^2}{875862} \\ &\qquad +\frac{y\left( 1579185x^2+655241y^2-2352261y+1159614 \right) ^2}{30205299618}\\ &\qquad+\frac{x^2y(86x^2+79y-237)^2}{43} \\ \ge\,& 0. \end{align*}

We are done.