parmenides51 wrote: Find the smallest value that the expression takes $x^4 + y^4 - x^2y - xy^2$, for positive numbers $x$ and $y$ satisfying $x + y \le 1$. Let $x,y$ be positive numbers satisfying $x + y \le 1.$ Prove that $$x^4 + y^4 - x^2y - xy^2\geq -\frac{1}{8}$$Equality holds when $x=y=\frac{1}{2}.$ Let $x,y$ be positive numbers satisfying $x + y \le 1.$ Prove that $$x^4 + y^4 - kxy(x+y)\geq -\frac{2k-1}{8}$$Where $k\in N^+.$ Equality holds when $x=y=\frac{1}{2}.$

parmenides51 wrote: Find the smallest value that the expression takes $x^4 + y^4 - x^2y - xy^2$, for positive numbers $x$ and $y$ satisfying $x + y \le 1$. The answer is $\frac 18$. Equality holds when $x=y=\frac 12$. We will prove that $x^4+y^4-x^2y-xy^2+\frac 18\ge 0$. Since $x+y\leq 1$, we have $\frac 18\ge \frac {(x+y)^2}8$.Thus, it suffices to prove that $x^4+y^4+\frac {(x+y)^2}8\ge x^2y+xy^2$. By AM-GM, we have $$\underbrace{\frac{x^4}{16}+\cdots +\frac{x^4}{16}}_{10 \ \text{times}}+\underbrace{\frac{y^4}{16}+\cdots +\frac{y^4}{16}}_{6 \ \text{times}}+\underbrace{\frac{x^2}{64}+\cdots +\frac{x^2}{64}}_{8 \ \text{times}}+\underbrace{\frac{xy}{64}+\cdots +\frac{xy}{64}}_{8 \ \text{times}}\ge 32\sqrt[32]{x^{64}\cdot y^{32}\cdot 2^{-160}}=x^2y$$$$\underbrace{\frac{y^4}{16}+\cdots +\frac{y^4}{16}}_{10 \ \text{times}}+\underbrace{\frac{x^4}{16}+\cdots +\frac{x^4}{16}}_{6 \ \text{times}}+\underbrace{\frac{y^2}{64}+\cdots +\frac{y^2}{64}}_{8 \ \text{times}}+\underbrace{\frac{xy}{64}+\cdots +\frac{xy}{64}}_{8 \ \text{times}}\ge 32\sqrt[32]{y^{64}\cdot x^{32}\cdot 2^{-160}}=y^2x$$Summing up these inequalities gives us the desired result.

beautifull solution!

Let $x,y$ be positive numbers satisfying $x + y \le 1.$Prove that $$xy+\frac{1}{x^2}+\frac{1}{y^2} \geq \frac{33}{4}$$

sqing wrote: Let $x,y$ be positive numbers satisfying $x + y \le 1.$ Prove that $$x^4 + y^4 - x^2y - xy^2\geq -\frac{1}{8}$$Equality holds when $x=y=\frac{1}{2}.$ Solution of Zhangyanzong: $$x^4 + y^4 - x^2y - xy^2\geq 2x^2y^2-xy=2\left(xy-\frac{1}{4}\right)^2 -\frac{1}{8} \geq -\frac{1}{8}$$

I think anwer is -1,25 . Because, x⁴+y⁴-x²y-y²x>= xy-xy(x+y)-0,625-0,625>=xy-xy-1,25=-1,25 Equality occurs when x=y=0,5.