1. If we let $x=y=\frac{\sqrt{2}}{2}$, the inequality becomes $\frac{1}{2}+\frac{1}{2}+1\geq C\cdot\sqrt{2}$ or $C\leq\sqrt{2}$. But if $C=\sqrt{2}$, then we can see, through expanding, that $x^2+y^2+1=\left(x-\frac{\sqrt{2}}{2}\right)^2+\left(y-\frac{\sqrt{2}}{2}\right)^2+\sqrt{2}(x+y)\geq\sqrt{2}(x+y)$. Hence $\boxed{\sqrt{2}}$ is the maximal value for $C$. 2. If we let $x=y=\frac{\sqrt{3}}{3}$, the inequality becomes $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+1\geq C\cdot\frac{2\sqrt{3}}{3}$, or $C\leq\sqrt{3}$. But if $C=\sqrt{3}$, then we can see through expanding, that $x^2+xy+y^2+1=\frac{3}{4}\left(x^2+2xy+y^2-\frac{4\sqrt{3}}{3}y-\frac{4\sqrt{3}}{3}x+\frac{4}{3}\right)$ $+\frac{1}{4}(x^2-2xy+y^2)+\sqrt{3}(x+y)$ $=\frac{3}{4}\left(x+y-\frac{2\sqrt{3}}{3}\right)^2+\frac{1}{4}(x-y)^2+\sqrt{3}(x+y)\geq \sqrt{3}(x+y)$. Hence $\boxed{\sqrt{3}}$ is the maximal value for $C$.

Let $a$, $b$, $c$, $d$, $e$, $f$, and $\eta$ be real numbers such that $a \geq 0$, $c \geq 0$, $f \geq 0$, and $b^2-4ac \leq 0$. Apparently, we have \[ax^2+bxy+cy^2+\eta(dx+ey)+f \geq 0\] for all reals $x$ and $y$ if \[|\eta| \leq \sqrt{\frac{f\left|b^2-4ac\right|}{ae^2-bde+cd^2}}\,,\] given that $ae^2-bde+cd^2 > 0$. The right-hand side quantity is the maximum possible value of $|\eta|$.

PS: If $a<0$, $c<0$, $f<0$, or $b^2-4ac > 0$, then no such $\eta$ exists.

momo1729 wrote: DMFA MC Find the maximum value of the real constant $C$ such that : $\forall (x,y) \in \mathbb{R}^{2}\ x^{2}+y^{2}+1\geq C(x+y).$ $\forall (x,y) \in \mathbb{R}^{2}\ x^{2}+y^{2}+xy+1\geq C(x+y).$ \[\boxed{\boxed{x+y,\ x-y\in\mathbb{R}\Longleftrightarrow x,\ y\in\mathbb{R}}}\] Let $A=x+y,\ B=x-y$, it is sufficient to consider the case of $x\geq y>0$, or $A>0,\ B\geq 0$. For 1), $A^2+B^2+2\geq 2CA\Longleftrightarrow \frac 12A+\frac{1}{A}+\frac{B^2}{2A}\geq C$, thus $C_{max}$ can be given the minimum of the L.H.S. By A.M.-G.M., the L.H.S.$\geq 2\sqrt{\frac 12A\cdot \frac{1}{A}}=\boxed{\sqrt{2}}$, when $\frac12A=\frac{1}{A}\ (A>0)$ and $B=0\Longleftrightarrow A=\sqrt{2},\ B=0\Longleftrightarrow x=y=\frac {1}{\sqrt{2}}$. For 2), In the same way as 1), $2(A^2+B^2)+A^2-B^2+4\geq 4CA\Longleftrightarrow \frac 34A+\frac {1}{A}+\frac{B^2}{4A}\geq C$, thus the L.H.S.$=\frac 34A+\frac {1}{A}+\frac{B^2}{4A}\geq 2\sqrt{\frac34A\cdot \frac{1}{A}}=\boxed{\sqrt{3}}$, when $\frac 34A=\frac{1}{A}\ (A>0)$ and $B=0\Longleftrightarrow A=\frac{2}{\sqrt{3}},\ B=0\Longleftrightarrow x=y=\frac{1}{\sqrt{3}}.$

tenniskidperson3 wrote: 2. If we let $x=y=\frac{\sqrt{3}}{3}$, the inequality becomes ... How did you get the idea of setting $x=y=\frac{\sqrt{3}}{3}$ ?

What I did is I let $x=a+b$ and $y=a-b$ for some $a, b$. Then the equation turns into $(a+b)^2+(a+b)(a-b)+(a-b)^2+1\geq C((a+b)+(a-b))$ $a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2+1\geq2aC$ $3a^2+b^2+1\geq2aC$ So we can vary $b$, and the left hand side is the minimum when $b=0$. So we can let $b=0$ and we get $3a^2+1\geq2aC$. Then we can subtract $2\sqrt{3}a$ from both sides to get $3a^2-2\sqrt{3}a+1=(\sqrt{3}a-1)^2\geq a(2C-2\sqrt{3})$ Thus when $a=\frac{\sqrt{3}}{3}$, we need $0\geq a(2C-2\sqrt{3})$ or $C\leq\sqrt{3}$. When $a=\frac{\sqrt{3}}{3}$ and $b=0$, we get $x=y=\frac{\sqrt{3}}{3}$.

Thank you. You can also set $x=y$, isolate C and derive the expression on the other hand so as to find its max.

momo1729 wrote: Find the maximum value of the real constant $C$ such that $x^{2}+y^{2}+1\geq C(x+y)$, and $ x^{2}+y^{2}+xy+1\geq C(x+y)$ for all reals $x,y$. 1) \[x^2+y^2+1\geq C(x+y)\Longleftrightarrow \left(x-\frac{C}{2}\right)^2+\left(y-\frac{C}{2}\right)^2+1-\frac{C^2}{2}\geq 0,\ \forall{x,\ y\in{\mathbb R}}\] $\therefore 1-\frac{C^2}{2}\geq 0\Longleftrightarrow -\sqrt{2}\leq C\leq \sqrt{2}$ with equality $x=y=\frac{C}{2}=\frac{1}{\sqrt{2}}.$ 2) \[x^2+y^2+xy+1\geq C(x+y) \Longleftrightarrow \left(x+\frac{y-C}{2}\right)^2+\frac34\left(y-\frac{C}{3}\right)^2+1-\frac{C^2}{3}\geq 0, \ \forall{x,\ y\in{\mathbb R}}\] $\therefore 1-\frac{C^2}{3}\geq 0\Longleftrightarrow -\sqrt{3}\leq C\leq \sqrt{3}$ with the equality ${x+\frac{y-C}{2}=0,\ y=\frac{C}{3}\Longleftrightarrow x=y=\frac{1}{\sqrt{3}}}.$

Thank you ! What do you think of using differentiation ?