From problem we know: $$(x+y)(x^2+y^2)=1$$Then we need to prove: $$\frac{4}{3}(x^6-y^6)\ge(x-y)(x^2+y^2)$$Then: $$4(x^2+y^2-xy)(x^3+y^3)\ge3(x^2+y^2)$$$$4((x^2+y^2)^2-(xy)^2)(x+y)\ge3(x^2+y^2)$$$$4((x^2+y^2)^2-(xy)^2)\ge3(x^2+y^2)^2$$Then by AM-GM: $$(x^2+y^2)^2\ge4(xy)^2$$So we're done....!

andria wrote: $x,y>0$ are two real numbers such that $x^4-y^4=x-y$. prove that: $$\frac{x-y}{x^6-y^6}\leq \frac{4}{3}(x+y)$$ This problem should be in middle school math... it's obvious for every body...

Another perspective is that we can use the condition to homogenize it. That is $$(\frac{x^4-y^4}{x-y})^2\cdot \frac{x-y}{x^6-y^6}\le \frac43(x+y)$$The rest is simple calculations.

andria wrote: $x,y>0$ are two real numbers such that $x^4-y^4=x-y$. prove that: $$\frac{x-y}{x^6-y^6}\leq \frac{4}{3}(x+y)$$ $x^4-y^4=x-y\Longleftrightarrow (x-y)(x+y)(x^2+y^2)=x-y$, since $x-y\neq 0$, we get $(x+y)(x^2+y^2)=1$, thus, we have $\frac{(x^2+y^2)(x^4-y^4)}{x^6-y^6}=\frac{x^6-y^6+x^2y^2(x^2-y^2)}{x^6-y^6}$ $=1+\frac{x^2y^2(x^2-y^2)}{x^6-y^6}=1+\frac{x^2y^2}{x^4+x^2y^2+y^4}\le \frac 43\ (\because x^4+y^4\ge 2x^2y^2).$ Therefore, we obtain $\frac{(x^2+y^2)(x^4-y^4)}{x^6-y^6}\le \frac 43=\frac 43(x+y)(x^2+y^2)$, yielding $\frac{x^4-y^4}{x^6-y^6}\le \frac 43\ (x+y)$, or $\frac{x-y}{x^6-y^6}\leq \frac{4}{3}(x+y).$

Remember that any homogeneous expression of order zero can always be viewed as a function of the variables ratio.

Let $x+y:=s$ and $xy:=p$ so that the given condition yields, (using the obvious $(\sqrt{x}-\sqrt{y})^2$), $$1=s(s^2-2p) \geq s(s^2-s^2/2) = \frac{s^3}{2} \iff s \leq \sqrt[3]{2} ...(1)$$. Also the given inequality is equivalent to $$\frac{3}{4} \leq (x+y)(x^5+x^4y+x^3y^3+xy^4+y^5) = s(s^5-4ps^3+2p^2s+p^3) = s^7-4s^6-3s^4+8s^3+3s+4-\frac{1}{s^2}$$which follows from $(1)$ using standard methods of calculus.

andria wrote: $x,y>0$ are two real numbers such that $x^4-y^4=x-y$. prove that: $$\frac{x-y}{x^6-y^6}\leq \frac{4}{3}(x+y)$$

andria wrote: $x,y>0$ are two real numbers such that $x^4-y^4=x-y$. prove that: $$\frac{x-y}{x^6-y^6}\leq \frac{4}{3}(x+y)$$ The hypothesis gives $$\frac{x-y}{x^6-y^6}=\frac{(x-y)^2}{(x^6-y^6)(x-y)}=\frac{(x^4-y^4)^2}{(x-y)(x^6-y^6)}$$Then $$\frac{4}{3}(x+y)-\frac{(x^4-y^4)^2}{(x-y)(x^6-y^6)}=\frac{4(x+y)(x-y)(x^6-y^6)-3(x^4-y^4)^2}{3(x-y)(x^6-y^6)}=\frac{(x^2-y^2)^4}{3(x-y)(x^6-y^6)} \geq 0$$And the result follows. $\square$

Let $x,y$ be two positive real numbers such that $x^4-y^4=x-y$. Prove that $$\frac{x-y}{x^8-y^8}< 2\sqrt[3]{2}(x+y)$$$$\frac{x-y}{x^8-y^8}<4(x^2+y^2)$$$$\frac{x^2-y^2}{x^6-y^6}< \frac{4\sqrt[3]{2}}{3}(x+y).$$Let $x,y$ be two positive real numbers such that $x^2-y^2=x-y$. Prove that $$\frac{x-y}{x^6-y^6}< \frac{16}{3}$$Let $x,y$ be two positive real numbers such that $x^3-y^3=x-y$ and $xy>0. $ Prove that $$\frac{x-y}{x^6-y^6}< \frac{9}{4}(x+y)$$(SXTX,11(2020))

$\frac{x^4-y^4}{x^6-y^6} = \frac{(x+y)(x-y)(x^2+y^2)}{(x^2+y^2)(x^4-x^2y^2+y^4)} = \frac{(x-y)(x+y)}{(x^4-x^2y^2+y^4)} \leq \frac{4}{3}(x+y)$ Then we need to prove : $\frac{x-y}{x^4-x^2y^2+y^4} = \frac{x^4-y^4}{x^4-x^2y^2+y^4} \leq \frac{4}{3}$ $3x^4-3y^4 \leq 4x^4+4y^4-4x^2y^2 $ or we need prove $0 \leq x^4+7y^4-4x^2y^2 = (x^2-2y^2)^2+3y^4 $ $\square$