parmenides51 wrote: If $x$ and $y$ are nonnegative real numbers with $x+y= 2$, show that $x^2y^2(x^2+y^2)\le 2$. $$ xy\leq 1$$$$ x^2y^2(x^2+y^2)=x^2y^2((x+y)^2-2xy)=2x^2y^2(2-xy)\leq2xy(2-xy)\leq 2\left(\frac{xy+2-xy}{2}\right)^2=2.$$

Let $a$ and $b$ be positive real numbers such that $a+b=2;$ $k$ is a given constant. Find the smallest $n=n(k)$ such that $$(ab)^n(a^k+b^k) \le 2.$$Let $a, b$ be positive real numbers satisfying $a+b =2.$ Show that$$a^3b^3(a^2+b^2)^2\le 4.$$

Lex $x,y\geq 0$ and $x+y= 2.$ Show that $$(x+1)y(x^2+y^2)\le 8$$$$(x+1)(y+1)(x^2+y^2)\le 12$$$$(x+1)(y-1)(x^2+y^2)\le 4$$$$(x+1)^2(y+1)^2(x^2+y^2)\le \frac{1000}{27}$$$$(x+1)^2(y-1)^2(x^2+y^2)\le 36$$Lex $x,y\geq 0$ and $x+2y= 2.$ Show that$$x^2y^2(x^2+y^2)\le \frac{16(20+2\sqrt[3]{10}+5\sqrt[3]{100})}{2025}$$

parmenides51 wrote: If $x$ and $y$ are nonnegative real numbers with $x+y= 2$, show that $$x^2y^2(x^2+y^2)\le 2$$ Irish 2000 Let $x$, $y$ are positive reals such that $x + y = 2$ , show that $$xy(x^2+y^2)\le 2$$ sqing wrote: Lex $x,y\geq 0$ and $x+y= 2.$ Show that $$(x+1)y(x^2+y^2)\le 8$$

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parmenides51 wrote: If $x$ and $y$ are nonnegative real numbers with $x+y= 2$, show that $x^2y^2(x^2+y^2)\le 2$. If $x$ or $y$ is $0$, then the inequality clearly holds. Henceforth, we assume $x, y > 0$. By AM-GM, we know $\sqrt{xy} \le 1$. Thus, $$x^3y^3 + 1 \ge 2 \sqrt{x^3y^3} \ge \left(2 \sqrt{x^3y^3} \right) \cdot \sqrt{xy} = 2x^2y^2.$$Now, observe $$1 \ge 2x^2y^2 - x^3y^3 = x^2y^2(2 - xy)$$which implies $$2 \ge x^2y^2(4 - 2xy) = x^2y^2((x+y)^2 - 2xy) = x^2y^2(x^2 + y^2)$$as desired. Equality holds when $x, y = 1$. $\blacksquare$

sqing wrote: Lex $x,y\geq 0$ and $x+y= 2.$ Show that $$(x+1)y(x^2+y^2)\le 8$$$$(x+1)(y+1)(x^2+y^2)\le 12$$$$(x+1)(y-1)(x^2+y^2)\le 4$$$$(x+1)^2(y+1)^2(x^2+y^2)\le \frac{1000}{27}$$$$(x+1)^2(y-1)^2(x^2+y^2)\le 36$$Lex $x,y\geq 0$ and $x+2y= 2.$ Show that$$x^2y^2(x^2+y^2)\le \frac{16(20+2\sqrt[3]{10}+5\sqrt[3]{100})}{2025}$$

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