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What is the maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$?
Wolfram Alpha plots this expression on a 3d graph, but I want to solve it algebraicly, by modifying the expression
My Attempts
1) add and substract 2 at the equation and we get $\frac{x+y+1}{y+1}+\frac{x+y+1}{x+1}$ and the numerator is same
=>failed
2) use AM-GM or Cauchy-Schwarz inequality
=>also failed

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    $\begingroup$ Since $x$ and $y$ both vary from $0$ to $1$ you can use one variable calculus. First maximize over $x$ and then over $y$. $\endgroup$ Dec 14, 2019 at 12:54
  • $\begingroup$ The function is convex in both variables, so that the maximum is attained at one of the corners of the quadrilateral. $\endgroup$
    – Martin R
    Dec 14, 2019 at 13:07

4 Answers 4

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When $ 0\le x,y \le 1$, then $$\frac{x}{y+1} \le \frac{x}{x+y}$$ $$\frac{y}{x+1} \le \frac{y}{x+y}$$ Adding them we have $$\frac{x}{y+1}+\frac{y}{x+1} \le \frac{x+y}{x+y}=1.$$

Equality holds when $x=0$ and $y=1$ or $x=1$ and $y=0$ and the maximum of $1$ is attained.

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It is $$\frac{x}{y+1}+\frac{y}{x+1}\le 1$$ since $$x(x+1)+y(y+1)\le (x+1)(y+1)$$ This is equivalent to $$0\le x(1-x)+(1-y)(1-x+y)$$

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    $\begingroup$ Why is $ x(x+1)+y(y+1)\le (x+1)(y+1)$? $\endgroup$
    – dissolve
    Dec 14, 2019 at 12:46
  • $\begingroup$ $\frac{x}{\color{#25f}{y+1}}+\frac{y}{\color{#d05}{x+1}}\le 1 \implies x\color{#d05}{(x+1)}+y\color{#25f}{(y+1)}\le\color{#d05}{(x+1)}\color{#25f}{(y+1)}$ $\endgroup$ Dec 14, 2019 at 12:50
  • $\begingroup$ You must say that value $1$ is actually obtained for $x=y=1$. $\endgroup$
    – Jean Marie
    Dec 14, 2019 at 13:04
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Given the function $f : \mathcal{D} \to \mathbb{R}$ of law:

$$ f(x,\,y) := \frac{x}{y + 1} + \frac{y}{x + 1}\,, $$

with $\mathcal{D} := [0,\,1] \times [0,\,1]$, since:

$$ \nabla f(x,\,y) = \left(\frac{1}{y + 1} - \frac{y}{\left(x + 1\right)^2}, \; \frac{1}{x + 1} - \frac{x}{\left(y + 1\right)^2}\right) \ne (0,\,0) \; \; \forall \, (x,\,y) \in D $$

it follows that f has no critical points inside $\mathcal{D}$.

So, studying $f$ on the boundary of $\mathcal{D}$, noting that:

$$ f'(0,\,t) = f'(t,\,0) = 1\,, \; \; \; f'(1,\,t) = f'(t,\,1) = \frac{1}{2} - \frac{1}{(t + 1)^2} $$

with $0 < t < 1$, $f$ has two critical points of coordinates $\left(1,\,\sqrt{2}-1\right)$, $\left(\sqrt{2}-1, \; 1\right)$, to which must be added the critical points placed in the four vertices of $\mathcal{D}$: $(0,\,0)$, $(1,\,0)$, $(1,\,1)$, $(0,\,1)$. This done, noting that:

$$ f\left(1,\,\sqrt{2}-1\right) = f\left(\sqrt{2}-1,\,1\right) = \sqrt{2} - \frac{1}{2}, \\ f(0,\,0) = 0, \; \; f(1,\,0) = f(1,\,1) = f(0,\,1) = 1 $$

since $f$ is a continuous function in $\mathcal{D}$, set closed and bounded, by Weierstrass's theorem:

$$ \underset{\mathcal{D}}{\min} f = 0 \; \; \; \; \; \; \underset{\mathcal{D}}{\max} f = 1 $$

as easily verifiable also in Wolfram Mathematica by the following code:

DensityPlot[x/(y + 1) + y/(x + 1), {x, 0, 1}, {y, 0, 1}, 
            FrameLabel -> {x, y}, PlotLegends -> Automatic]

enter image description here

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Since $$ \delta\left(\frac{x}{y+1}+\frac{y}{x+1}\right) =\left(\frac1{y+1}-\frac{y}{(x+1)^2}\right)\delta x +\left(\frac1{x+1}-\frac{x}{(y+1)^2}\right)\delta y $$ to get an interior critical point, we need $$ \frac1{y+1}=\frac{y}{(x+1)^2}\quad\text{and}\quad\frac1{x+1}=\frac{x}{(y+1)^2} $$ which have only the singular solution $x=y=-1$, which, even if allowed, are outside of $[0,1]\times[0,1]$. Therefore, we are searching for boundary critical points.

Due to the symmetry, we need only consider $x=0$ and $x=1$.

For $x=0$, we get $\frac{x}{y+1}+\frac{y}{x+1}=y$ and the maximum is at $y=1$.

For $x=1$, we get $\frac{x}{y+1}+\frac{y}{x+1}=\frac1{y+1}+\frac y2$. Then, the critical point is $y=\sqrt2-1$, which gives $\frac{x}{y+1}+\frac{y}{x+1}=\sqrt2-\frac12\lt1$. The endpoints $y\in\{0,1\}$ both give $\frac{x}{y+1}+\frac{y}{x+1}=1$.

Thus, the maximum is $1$.

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