${{x}^{2}}+{{y}^{2}}+1\ge 2(xy-x+y)\Leftrightarrow {{(x-y)}^{2}}+2(x-y)+1\ge 0\Leftrightarrow {{(x-y+1)}^{2}}\ge 0$ equality when x-y+1=0.

Prove that the inequality $$x^2+y^2+16\ge xy+4\sqrt {3}x-4\sqrt 3 y$$$$x^2+y^2+6\ge xy-3\sqrt {2}x+3\sqrt {2} y$$is satisfied by any $x$, $y$ real numbers. Indicate when the equality is satisfied.

Let $x,y\geq 0 .$ Prove that $$x^2+y^2+12\ge xy-4\sqrt {3x}+4\sqrt 3\\\ y$$$$x^2+y^2+\frac{9}{2}\ge xy-3\sqrt {2x}+3\sqrt {2}\\\ y$$$$x^2+y^2+\frac{3}{2\sqrt[3]{4}}\ge xy+2\sqrt {2x}-3\sqrt{3} \\\ y$$$$x^2+y^2+\frac{3\sqrt[3]{9}}{2\sqrt[3]{2}}\ge xy+2\sqrt {3x}-4\sqrt{3} \\\ y$$

sqing wrote: Prove that the inequality $$x^2+y^2+16\ge xy+4\sqrt {3}x-4\sqrt 3 y$$$$x^2+y^2+6\ge xy-3\sqrt {2}x+3\sqrt {2} y$$is satisfied by any $x$, $y$ real numbers. Indicate when the equality is satisfied. a) ${{x}^{2}}+{{y}^{2}}+16\ge xy+4\sqrt{3}x-4\sqrt{3}y\Leftrightarrow {{x}^{2}}-(y+4\sqrt{3})x+{{y}^{2}}+4\sqrt{3}y+16\ge 0$ ${{\Delta }_{x}}=-3{{y}^{2}}-8\sqrt{3}y-64=-{{(y+\sqrt{3}y)}^{2}}-48<0$ b) ${{x}^{2}}+{{y}^{2}}+6\ge xy-3\sqrt{2}x+3\sqrt{2}y\Leftrightarrow {{x}^{2}}-(y-3\sqrt{2})x+{{y}^{2}}-3\sqrt{2}y+6\ge 0$ ${{\Delta }_{x}}=-3{{y}^{2}}+6\sqrt{2}y-6=-3{{(y-\sqrt{2})}^{2}}\le 0$

continue #5

#5 a) ${{x}^{2}}+{{y}^{2}}+12\ge xy-4\sqrt{3x}+4\sqrt{3}y\Leftrightarrow f(x)={{x}^{2}}-(y-4\sqrt{3})x+{{(y-2\sqrt{3})}^{2}}\ge 0$ and $f(0)\ge 0$

@sqing, how do you obtain those constants?

The last inequality it is just $(x-y+1)^2>=0$

parmenides51 wrote: Prove that the inequality $x^2+y^2+1\ge 2(xy-x+y)$ is satisfied by any $x$, $y$ real numbers. Indicate when the equality is satisfied.

sqing wrote: Let $x,y\geq 0 .$ Prove that $$x^2+y^2+12\ge xy-4\sqrt {3x}+4\sqrt 3\\\ y$$