# Michael Rozenberg's Inequality in Two Variables

### Problem

### Solution 1

Due to the constraint $x+y=2,\,$ $xy\in [0,1].\,$ Let $xy=p.\,$ Then

$\displaystyle\sqrt{x^2+3}+\sqrt{y^2+3}=\sqrt{10-2p+2\sqrt{(p-3)^2+12}}$

so that the required inequality becomes

$\displaystyle\sqrt{10-2p+2\sqrt{(p-3)^2+12}}\ge 6-\sqrt{p+3}.$

Squaring gives

$\displaystyle 2\left(\sqrt{(p-3)^2+12}+6\sqrt{p+3}\right)-3p\ge 29.$

Let $f(p)\,$ denote the left-hand side. We have

$\displaystyle f'(p)=2\left(\frac{p-3}{\sqrt{(p-3)^2+12}}+\frac{3}{\sqrt{p+3}}\right)-3$

and

$\displaystyle\begin{align} f''(p)&=6\left(\frac{4}{[(p-3)^2+12]^{3/2}}-\frac{1}{2(p+3)^{3/2}}\right)\\ &=6\left(\frac{\displaystyle 1}{\displaystyle\left[\frac{(p-3)^2+12}{4^{2/3}}\right]^{3/2}}-\frac{1}{[2^{2/3}(p+3)]^{3/2}}\right) \end{align}$

We shall prove that $f''(0)\le 0,\,$ for $p\in [0,1].\,$ To this end, suffice it to show that, for $p\in [0,1],$

$\displaystyle \left[\frac{(p-3)^2+12}{4^{2/3}}\right]^{3/2}\ge [2^{2/3}(p+3)]^{3/2},$

or equivalently

$\displaystyle (p-3)^2+12-4(p+3)\ge 0,$

which is $(p-1)(p-9)\ge 0\,$ and obviously holds for $p\in [0,1].\,$ Thus it follows that $\max_{p\in [0,1]}f'(p)=f'(0)\lt 0,\,$ implying $\min_{p\in [0,1]}f(p)=f(1)=29,\,$ which completes the proof.

### Solution 2

Define, for $x\in[0,2],$

$f(x)=\sqrt{x^2+3}+\sqrt{x^2-4x+7}+\sqrt{-x^2+2x+3}.$

Then

$\displaystyle f'(x)=\frac{x}{\sqrt{x^2+3}}+\frac{x-2}{\sqrt{x^2-4x+7}}+\frac{1-x}{\sqrt{-x^2+2x+3}}.$

$f'=0\,$ for $x=1.\,$ Hence, $2\sqrt{3}+\sqrt{7}\ge f\ge6.$

By convexity, $x=y:$

### Illustration

### Acknowledgment

Leo Giugiuc has kindly communicated to me the above problem by Michael Rozenberg (Israel), originally posted at the artofproblemsolving forum, along with his solution.

The illustration is by Gary Davis.

### Inequalities with the Sum of Variables as a Constraint

- An Inequality with Constraint
- An Inequality with Constraints II
- An Inequality with Constraint V
- An Inequality with Constraint VI
- An Inequality with Constraint XI
- Monthly Problem 11199
- Problem 11804 from the AMM
- Sladjan Stankovik's Inequality With Constraint
- Sladjan Stankovik's Inequality With Constraint II
- An Inequality with Constraint V
- An Inequality with Constraint VI
- An Inequality with Constraint XI
- An Inequality with Constraint XII
- An Inequality with Constraint XIII
- Inequalities with Constraint XV and XVI
- An Inequality with Constraint XVII
- An Inequality with Constraint in Four Variables
- An Inequality with Constraint in Four Variables II
- An Inequality with Constraint in Four Variables III
- An Inequality with Constraint in Four Variables IV
- Inequality with Constraint from Dan Sitaru's Math Phenomenon
- An Inequality with a Parameter and a Constraint
- Cyclic Inequality with Square Roots And Absolute Values
- From Six Variables to Four - It's All the Same
- Michael Rozenberg's Inequality in Three Variables with Constraints
- Michael Rozenberg's Inequality in Two Variables
- Dan Sitaru's Cyclic Inequality in Three Variables II
- Dan Sitaru's Cyclic Inequality in Three Variables IV
- Dan Sitaru's Cyclic Inequality in Three Variables VI
- An Inequality with Arbitrary Roots
- Inequality 101 from the Cyclic Inequalities Marathon
- Sladjan Stankovik's Inequality With Constraint II
- An Inequality with Constraint in Four Variables
- An Inequality with Constraint in Four Variables IV
- Cyclic Inequality with Square Roots And Absolute Values
- From Six Variables to Four - It's All the Same
- Michael Rozenberg's Inequality in Two Variables
- Dan Sitaru's Cyclic Inequality in Three Variables II
- Inequality 101 from the Cyclic Inequalities Marathon

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