# Inequality, finding Constant

Can anyone help me with this Math Olympiad Task from 2007 from Germany? I want to find the smallest $$C$$, such that for every $$x,y \in \mathbb{R}$$ the inequality: $$1+(x+y)^2 \leq C(1+x^2)(1+y^2)$$ holds. I know that I have to maximize the function $$a(x, y) = \frac{1+(x+y)^2}{(1+x^2)(1+y^2)}$$ which gives me $$\frac{4}{3}.$$

P.S.: Thanks for the help y'all :)

• (Side remark: This is Problem 471344.) Jun 24, 2021 at 19:37
• Since you claim to have solved this, can you add your solution? Jun 24, 2021 at 22:36

Since it must be true for all $$x,y$$ then it must be true also for $$x=y$$ and we get $$Ct^2-4t+3\geq 0$$ where $$t=x^2+1\geq 1$$.

Clearly $$C>0$$ so we can rewrite it like this $$(Ct-2)^2 -4+3C\geq 0$$

If $$C<4/3$$ then for $$t={2\over c} >1$$

we get $$-4+3C\geq 0 \implies C\geq {4\over 3}$$

Now, the game is not over yet. You must prove that this value actually works for all $$x,y$$:

$$3+3x^2+6xy+3y^2\leq 4+4x^2+4y^2+4x^2y^2$$ which is equivalent to $$0\leq (x-y)^2+(2xy-1)^2$$ which is obviously true.

• Yes, this works now. Deleted my comments. Jun 25, 2021 at 15:52
• Yeah you edited your answer now so it looks fine to me Jun 26, 2021 at 14:05

Simple Verification

Once we know the constant is $$\frac43$$, the inequality is fairly simple. $$1+(x+y)^2\le\frac43\left(1+x^2\right)\left(1+y^2\right)\tag1$$ is equivalent to $$\color{#C00}{1+4x^2y^2}+\color{#090}{x^2+y^2}\ge6xy\tag2$$ which is true because the AM-GM says $$\color{#C00}{1+4x^2y^2}\ge4xy$$ and $$\color{#090}{x^2+y^2}\ge2xy$$. Equality is attained when $$x=y=\frac1{\sqrt2}$$.

Variational Argument To Get The Constant

Suppose we have have $$\left(1+x^2\right)\left(1+y^2\right)$$ fixed and we wish to maximize $$1+(x+y)^2$$. That is, for all $$\delta x,\delta y$$ so that $$2x\left(1+y^2\right)\delta x+2y\left(1+x^2\right)\delta y=0\tag3$$ we also have $$2(x+y)(\delta x+\delta y)=0\tag4$$ $$(3)$$, $$(4)$$, and orthogonality implies that there is a $$\lambda$$ so that $$2(x+y)=\lambda2x\left(1+y^2\right)\qquad\text{and}\qquad2(x+y)=\lambda2y\left(1+x^2\right)\tag5$$ which implies $$x+\frac1x=y+\frac1y\tag6$$ which means that $$y=x\qquad\text{or}\qquad y=\frac1x\tag7$$ Let $$\boldsymbol{y=\frac1x}$$ \begin{align} \frac{1+(x+y)^2}{\left(1+x^2\right)\left(1+y^2\right)} &=\frac{1+\left(x+\frac1x\right)^2}{\left(1+x^2\right)\left(1+\frac1{x^2}\right)}\\ &=\frac{1+\left(x+\frac1x\right)^2}{\left(x+\frac1x\right)^2}\\ &\le\frac54\tag8 \end{align} since $$x+\frac1x\ge2$$ (equality when $$x=y=1$$).

Let $$\boldsymbol{x=y}$$ \begin{align} \frac{1+(x+y)^2}{\left(1+x^2\right)\left(1+y^2\right)} &=\frac{1+4x^2}{1+2x^2+x^4}\\ &=16\frac{1+4x^2}{16+32x^2+16x^4}\\[6pt] &=\frac{16}3\frac1{\frac{1+4x^2}3+2+\frac3{1+4x^2}}\\ &\le\frac43\tag9 \end{align} since $$\frac{1+4x^2}3+\frac3{1+4x^2}\ge2$$ (equality when $$x=y=\frac1{\sqrt2}$$).

Therefore, we get that $$1+(x+y)^2\le\frac43\left(1+x^2\right)\left(1+y^2\right)\tag{10}$$

The command of Mathematica

Resolve[ForAll[{x, y}, 1 + (x + y)^2 <= c*(1 + x^2)*(1 + y^2)], Reals]


answers $$c\geq \frac{4}{3}$$

Addition. and the command of Mathematica

Maximize[(1 + (x + y)^2)/(1 + x^2)/(1 + y^2), {x, y}]


$$\left\{\frac{4}{3},\left\{x\to -\frac{1}{\sqrt{2}},y\to -\frac{1}{\sqrt{2}}\right\}\right\}$$

shows how to derive it by hand.

• Oh, while I was editing my answer you updated yours too. Thank you anyway! Jun 24, 2021 at 19:47

Claim: We will show that $$1 + (x+y)^2 \leq \frac{4}{3} ( 1+x^2)(1+y^2)$$.

Proof: By expanding, WTS

$$4x^2y^2 + x^2 + y^2 + 1 \geq 6 xy.$$

This is true by applying AM-GM creatively:

$$4x^2 y^2 + 1 \geq 4 |xy | \geq 4 xy$$.
$$x^2 + y^2 \geq 2 | xy | \geq 2xy$$.

Equality holds iff $$2xy = 1$$, $$x = y$$ and $$xy \geq 0$$, which gives the solution set $$x = y = \pm \frac{1}{ \sqrt{2} }$$.
This solution sets also what that $$\frac{4}{3}$$ is the smallest possible value of $$C$$.

Note:

• As to how one can guess the value of $$C$$, we use the huge wishful thinking simplification that $$x=y$$, and have the quadratic equation in $$t = x^2$$ of
$$C t ^2 + ( 2 C - 4 )t + ( C - 1 ) \geq 0 \quad t \geq 0.$$ To satisfy this, we require
A) If $$t = - \frac{ 2C-4}{C } \geq 0$$, then $$f(t) \geq 0$$ $$\Rightarrow$$ if $$0 < C < 2$$, then $$C \leq \frac{4}{3}$$, so the solution set is $$2 > C \geq \frac{4}{3}$$.
B) Else if $$t = - \frac{ 2C-4}{C } \leq 0$$, then $$f(0) \geq 0$$ $$\Rightarrow$$ If $$t < 0$$ or $$t > 2$$, then $$C > 1$$. So the solution set is $$C \geq 2$$.
Hence, $$C \geq \frac{4}{3}$$, so the minimum value to try is $$C = \frac{4}{3}$$.
Note that this might not work because the equality condition might not occur at $$x = y$$.
• @Aqua Thanks for editing. I was responding to Joshua. Jun 25, 2021 at 15:55

Similar to @Acqua, I got it in this way:

Since it has to satisfy $$\forall x,y \in {\Bbb R}$$, it must satisfy with $$x=y=0$$. So this gives $$C(1+0)(1+0) \geqslant 1+0 \iff C \geqslant 1$$.

Also if $$x=y$$, then $$C(1+x^{2})^{2} \geqslant 1 + (2x)^2 \iff C(1+x^{2})^{2} \geqslant 1 + 4x^2 \iff C(1+x^{2})^{2} - 4x^2 - 1 \geqslant 0 \iff C(1+x^{2})^{2} - 4(x^2 + 1) + 3 \geqslant 0$$. Substituting $$z = x^2 + 1 \geqslant 1, \forall x \in \Bbb R$$, and since it is true $$\forall z \in \Bbb R$$ such that $$z \geqslant 1$$, implies that $$Cz^2 -4z +3 \geqslant 0$$ is also true.

Notice that this is a quadratic equation, and if we solve for $$z$$, we have: $$z = \frac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(C)(3)} }}{{2C}} = \frac{{4 \pm \sqrt {16 - 12C} }}{{2C}}$$ Since $$C \geqslant 1$$ and $$z \in \Bbb R$$ such that $$z \geqslant 1 \Rightarrow$$ the discriminant $$d$$ of the quadratic inequality is either $$0$$ or less than $$0$$ (because $$16 - 12C \leqslant 0$$ for some $$C \in \Bbb R$$). If $$d$$ (discriminant of the quadratic equation) is less than $$0$$, $$z$$ has complex roots. Therefore, $$d=0 \Rightarrow 16 - 12C = 0 \Rightarrow C = \frac{4}{3}$$.

## Proof

If $$C= \frac {4}{3}$$, this implies that $$\frac{4}{3}(1+x^2)(1+y^2) \geqslant 1 + (x+y)^2$$. Using some algebra,

$$\frac{4}{3}(1+x^{2}y^{2}+x^2+y^2) \geqslant 1 + x^2 + 2xy + y^2$$

$$\frac{4}{3} + \frac{4}{3}{x^2} + \frac{4}{3}{y^2} + \frac{4}{3}{x^2}{y^2} - {x^2} - {y^2} - 2xy - 1\geqslant 0$$

$$\frac{1}{3} + \frac{1}{3}{x^2} + \frac{1}{3}{y^2} + \frac{4}{3}{x^2}{y^2} - 2xy \geqslant 0$$

$$\frac{1}{3} + \frac{1}{3}{x^2} + \frac{1}{3}{y^2} + \frac{4}{3}{x^2}{y^2} \geqslant 2xy$$

LHS is always positive and the RHS may be negative. Multiplying everything by 3, we have:

$$1 + x^2 + y^2 + 4x^{2}y^{2} \geqslant 6xy$$

$$1 + x^2 + y^2 + 2x^{2}y^{2} + 2x^{2}y^{2} \geqslant 6xy$$

$$1 + (x+y)^2 \geqslant 6xy - 2x^{2}y^{2}$$

Case 1:

If $$x=o$$ and $$y=\text{free} \Rightarrow 1 + y^2 \geqslant 0$$ (True)

Case2:

If $$y=o$$ and $$x=\text{free} \Rightarrow 1 + x^2 \geqslant 0$$ (True)

Case 3:

If $$x=y \Rightarrow 1 + 4x^2 \geqslant 1 \geqslant 0 \geqslant x^{2}(6-2x^2)$$(because $$x^2 \geqslant 0$$ and $$6 - 2x^2 \leqslant 0 \Rightarrow x^{2}(6-2x^2) \leqslant 0$$) (True)

Case 4:

If $$x=y=0 \Rightarrow 1 \geqslant 0$$ (true)

Case 5:

WLOG, if $$x = \text{fixed}$$ and $$y = \text{variable} \Rightarrow x=a= \text{constant}$$ and $$1 + (a+y)^2 \geqslant 6ay-2a^{2}y^{2}$$.

GRAPH: By looking at the graph, we see that this is always true.

ALGEBRAICALLY: $$1 + a^2 + 2ay - 6ay + y^2 + 2a^{2}y^{2} \geqslant 0$$ (moving things on the LHS). Then, when grouping: $$(1+a^2) -4ay + (1+2a^2)y^2 \geqslant 0$$. Let's check its roots:

\eqalign{ & y = \frac{{4a \pm \sqrt {16{a^2} - 4(1 + {a^2})(1 + 2{a^2})} }}{{2 + 4{a^2}}} \cr & = \frac{{4a \pm \sqrt { - 4 + 4{a^2} - 8{a^4}} }}{{2 + 4{a^2}}} \cr}

But, the discriminant is always less than $$0$$. This implies that the quadratic is actually greater than $$0$$, i.e. $$(1+a^2) -4ay + (1+2a^2)y^2 > 0$$

Q.E.D.

• Can you see my comments on Aqua's solution? I don't think this is correct. If I made an error, please explain. Jun 24, 2021 at 22:33
• In your comments, it is not assuming that the maximum of the quadratic "equation" is when $x=y$. Notice that the original problem says it must satisfy $\forall x,y \in \Bbb R$. Therefore, $C \in \Bbb R$ has to satisfy when $x=y$ case. Let me know if this helps, see ya!
– user938668
Jun 25, 2021 at 1:13
• Also, if $C=\frac{4}{3}$, the graph of $\frac{4}{3}z^2 -4z + 3$ is always nonnegative.
– user938668
Jun 25, 2021 at 1:31
• Addressing your solution specifically, 1) You have not shown that $\forall x, y \in \mathbb{R}$, $1 + ( x+y)^2 \leq 4/3 ( 1 + x^2 ) ( 1 + y^2)$. You have only attempted to show this n the special case of $x = y$. How does having it how in the special case imply the general case? 2) Are you sure that "$16 - 12 C \leq 0 \forall C \geq 1$? EG What about $C = 1$? 3) Just because the quadratic is $\geq 0$ in the domain $z \geq 1$, doesn't mean that the quadratic is always $\geq 0$. For example $z^2 - 1 \geq 0$ when $z \geq 1$, but not always. Jun 25, 2021 at 15:52
• Thanks for correcting me!
– user938668
Jun 25, 2021 at 17:57

$$\textbf{1. Solution.}$$ Let us find minimum $$C,$$ so that $$1+(x+y)^2 \leq C(1+x^2)(1+y^2). \tag{1}$$ We put into (1) the values $$x=y=1 / \sqrt{2}$$ and get $$3 \leq C\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2}\right)=\frac{9}{4} C$$ and thus necessarily $$C \geq 4 / 3$$. For all real numbers $$x, y$$ holds the inequality $$0 \leq(x-y)^{2}+(2 x y-1)^{2} .$$ Equivalent transformations yield $$6 x y \leq x^{2}+y^{2}+4 x^{2} y^{2}+1$$. From this follows further $$3+3 x^{2}+6 x y+3 y^{2} \leq 4+4 x^{2}+4 y^{2}+4 x^{2} y^{2}$$ and finally $$$$1+(x+y)^{2} \leq \frac{4}{3}\left(1+x^{2}\right)\left(1+y^{2}\right) \tag{2}$$$$ If $$C \geq 4 / 3$$ holds, then (1) follows immediately from (2). Thus, the smallest number $$C$$ we are looking for which satisfies (1) for all real numbers $$x$$ and $$y$$ is $$C=4 / 3$$.
Of course, the question arises how to arrive at such a solution, how to find the values $$x=y=1 / \sqrt{2}$$ or $$C=4 / 3$$. Here heuristic intuition helps, for example one can calculate through the special case $$x=y$$.

\textbf{2. solution.} We show that one can also solve the problem directly, systematically and stepwise as an extreme value problem. In doing so, we dispense with the differential calculus and only consider how to determine the minimum of a function $$y=x^{2}$$ for $$x \geq a$$. (Cf. Fig. 1.)

We obtain the unique minimum

$$$$x_{\min }=\left\{\begin{array}{lll} 0 & \text { falls } & a<0 \\ a & \text { falls } & a \geq 0 \end{array}, \quad \text { with the minimum } \quad y_{\min }=\left\{\begin{array}{lll} 0 & \text { falls } & a<0 \\ a^{2} & \text { falls } & a \geq 0 \end{array} .\right.\right. \tag{3}$$$$

To at least partially decouple the left-hand side in (1) from the right-hand side, we transform so that it depends on only one variable, $$x=s+t, \quad y=-s+t, \quad \text { so } \quad s=\frac{1}{2}(x-y), \quad t=\frac{1}{2}(x+y) .$$ This converts the left-hand side to $$1+(x+y)^{2}=1+4 t^{2}$$ and the inequality is satisfied exactly if it is correct for all $$t$$ and the minimum $$s=s_{\min }$$ of the right-hand side. We calculate this minimum (ignoring for now the factor $$C$$, which obviously must be positive). \begin{aligned} \left(1+x^{2}\right)\left(1+y^{2}\right) &=\left(1+s^{2}+t^{2}+2 s t\right)\left(1+s^{2}+t^{2}-2 s t\right) \\ &=\left(1+s^{2}+t^{2}\right)^{2}-4 s^{2} t^{2}=\left(1+s^{2}+t^{2}\right)^{2}-4\left(1+s^{2}\right) t^{2}+4 t^{2} \\ &=\left(1+s^{2}-t^{2}\right)^{2}+4 t^{2} \end{aligned} We can apply (3) with $$x=1+s^{2}-t^{2}$$ and $$a=1-t^{2}$$ and obtain $$\left(1+x^{2}\right)\left(1+y^{2}\right)=\left\{\begin{array}{ll} 4 t^{2} & \text { for } s^{2}=t^{2}-1>0, \\ \left(1-t^{2}\right)^{2}+4 t^{2}=\left(1+t^{2}\right)^{2} & \text { with } s^{2}=0, \text { for } t^{2} \leq 1 \end{array}\right.$$ So we distinguish the two cases $$t^{2} \leq 1$$ and $$t^{2}>1$$.

$$\textit{Case 1.}$$ For $$t^{2} \leq 1$$ is (1) exactly for $$1+4 t^{2} \leq C\left(1+t^{2}\right)^{2}$$ is satisfied. After division by $$C>0$$ this is equivalent to $$$$\begin{array}{r} t^{4}+2(1-2 / C) t^{2}+1-1 / C \geq 0.\\ \left(t^{2}+1-2 / C\right)^{2}+(3 C-4) / C^{2} \geq 0 \end{array} \tag{4}$$$$ To get the smallest value of the left-hand side, we can apply (3) again. We get the two cases $$\begin{array}{clrrr} 3 C-4 \geq 0, & \text { for the case } & t_{\min }^{2}=2 / C-1 \geq 0, & \text { i.e., if } & C \leq 2, \\ 1-1 / C \geq 0 & \text { or } \quad C \geq 1 & \text { and } \quad t_{\min }=0, & \text { if } & C>2. \end{array}$$ In summary, (4) is satisfied for all real $$t$$ exactly when $$C \geq 4 / 3$$. Since for $$C \geq 4 / 3$$ always $$t_{\min }^{2} \leq 2 \cdot 3 / 4-1=1 / 2<1$$ is satisfied, this equivalence of conditions is also valid under the constraint $$t^{2} \leq 1$$ is valid.

$$\textit{Case 2.}$$ For the other case, $$t^{2}>1$$, $$1+4 t^{2} \leq C \cdot 4 t^{2}$$ or $$4(C-1) t^{2} \geq 1$$ must be proved. First, $$C-1>0$$ must hold. The minimum of $$t^{2}$$ for $$t^{2} \geq 1$$ is $$t^{2}=1$$, so the estimate is satisfied for all $$t$$ of the second case exactly if $$4(C-1) \geq 1$$ holds, but this is always satisfied if the condition $$4 \leq 3 C$$ of the first case is already satisfied.

The inequality (1) is satisfied exactly for all $$x$$ and $$y$$, respectively for all real numbers $$s$$ and $$t$$, if $$C \geq 4 / 3$$ holds. The smallest constant we are looking for is $$C=4 / 3$$. $$\diamond$$