Two Maximized Constants in an Inequality

Let's work on the first equation first.

$x^2+y^2+1\ge C_1(x+y)$

Considering this as a quadratic wrt x we rearrange to get $x^2-C_1x+(y^2-C_1y+1)\ge 0$

Using the quadratic formula, we get $x=\dfrac{C_1\pm \sqrt{-4y^2+4C_1y+(C_1^2-4)}}{2}$

Note that if the discriminant $\Delta > 0$ , then there exists some $x$ that does not satisfy the inequality. Thus, we must have $\Delta \le 0$ , or $-4y^2+4C_1y+(C_1^2-4)\le 0$

Now we have to solve a quadratic wrt y. Using the quadratic formula again, we see that $y= \dfrac{C_1\pm \sqrt{2C_1^2-4}}{2}$

Similarly to above, note that if $\Delta > 0$ in this new quadratic, then there exists a $y$ such that it does not satisfy the inequality; thus, $\Delta \le 0$ , or $2C_1^2-4 \le 0$

Solving for $C_1$ , we get that $C_1\le \sqrt{2}$ , so the maximum value of $C_1$ is $\boxed{\sqrt{2}}$ .

This also means that $x=y = \dfrac{C_1}{2}$ is the equality case (since the discriminant in both cases disappears) so our equality case is $(x,y)=\left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)$

We have the equation $x^2+y^2+xy+1\ge C_2(x+y)$

Using a similar idea to before, we express the inequality as a quadratic wrt x: $x^2+(y-C_2)x+(y^2-C_2y+1)\ge 0$

Using the quadratic equation, we obtain $x=\dfrac{C_2-y\pm \sqrt{-3y^2+2C_2y+C_2^2-4}}{2}$

Similar to last time, we must have $\Delta \le 0$ , so $-3y^2+2C_2y+C_2^2-4\le 0$

Solving the new quadratic wrt y we have $y=\dfrac{2C_2\pm \sqrt{16C_2^2-48}}{6}$

Like before, we must have $\Delta \le 0$ , so $12C_2^2-48\le 0$

Solving for $C_2$ gives $C_2\le \sqrt{3}$ , so the maximum of $C_2$ is $\boxed{\sqrt{3}}$ .

This also means that the equality case is $y=\dfrac{C_2}{3}$ , and $x=\dfrac{C_2-y}{2}$ , which simplifies to $(x,y)=\left(\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{3}}{3}\right)$

Finishing of the problem, we have $C_1C_2=\sqrt{2}\cdot \sqrt{3}=\sqrt{6}$ , so our answer is $\boxed{6}$ .

I am pretty sure this was an extremely lucky guess. Now, I will post a solution, feel free to completely tear me apart for it.

Now, take $x^{2}+y^{2}+1$ . If $xy=\frac12$ , then the expression can be written as $(x+y)^{2}$

Now, for the maximum, equality must hold. Now, $(x+y)^{2}=C_1(x+y)\Rightarrow C_1=x+y$ . By A.M. G.M. Inequality, the minimum value for $x+y=\frac{2}{\sqrt{2}}=\sqrt{2}$ . Thus, $C_1=\sqrt{2}$ .

Now, $x^{2}+y^{2}+xy+1= ((x+y)^{2}+xy)$ because $xy$ is assumed to be $\frac12$ . Thus, the expression boils down to $C_2=\frac{5}{2\sqrt{2}}$ . Thus, $C_1C_2=\frac{5}{2}=\sqrt{\frac{25}{4}}\approx\sqrt{\boxed{6}}$

I am completely lost. This is a complete guess. Could someone else please post a solution?

For the first part, $x^2+y^2 + 1 \geq \dfrac{(x+y)^2}{2} + 1 \geq 2 \sqrt{\dfrac{(x+y)^2}{2}} = \sqrt{2}(x+y)$ . with equality when $x=y$ and when $(x+y)^2 = 2$ both of which simultaneouly indicate towards $(x,y) = \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}$ . For the second part, $x^2+y^2+xy+1 = (x+y)^2 + 1 - xy \geq (x+y)^2 +1 - \dfrac{(x+y)^2}{4} = \dfrac{3(x+y)^2}{4} + 1 \geq \sqrt{3}(x+y)$ . With equality only when $x=y$ and $(x+y) = \dfrac{2}{\sqrt{3}}$ hold simultaneously which hints towards $(x,y) = ( \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}})$ .

First part:

expressing in terms of perfect squares:

(x - sqrt(c1/2))^2 + (y - sqrt(c1/2))^2 + 1 - [(c1^2)/2] > 0

for c1 to be maximum and have this condition to be true

1 - [(c1^2)/2]=0

c1 = sqrt(2)

we have c1 = sqrt(2)

For the second part: Since all terms are symmetrical wrt x and y .. Lets re-express [x^2 + y^2 + xy + 1 - c2(x+y)] this in terms of:[As sum of squares]

(ax + ay + b)^2 + (1 - a^2)*x^2 + (1-a^2)y^2 + (1-2a^2)xy + (1-b^2) - (c2+ 2ab)(x+y)

We have 3 degrees of freedom here Constraints: if we can vanish 'xy' in the second part of the expression we can regroup with the 2 degrees of freedom left to regroup the remaining to perfect square.

Vanishing xy we get a = 1/sqrt(2);

we get for the rest of the expression, leaving the perfect square;

1/2* [x^2 + y^2 +2* (1-b^2) - 2 (c2+ b sqrt(2))(x+y)]

to make this a sum of perfect squares we have the condition

(c2 + b*sqrt(2))^2 = (1-b^2)

evaluating c2 in terms of b and then applying condition for maximum of c2 we end up with b = sqrt(2/3); condition c2 = 2ab + b/2a = sqt(3)..

interpretazione analitica: mettendo l'uguale si ottengono l'equazione di una circonferenza e di un'ellisse (ruotata) con il >= ho i punti ad esse esterne. Ho applicato la condizione affinché esse si riducessero ad un solo punto

Via Google translate: analytic interpretation: putting the same you will get the equation of a circle and an ellipse (rotated) with the> = I have the points to them outside. I applied the condition so that they were reduced to a single point