http://www.mathlinks.ro/Forum/viewtopic.php?t=16551

N.T.TUAN wrote: Suppose that positive numbers $x$ and $y$ satisfy $x^{3}+y^{4}\leq x^{2}+y^{3}$. Prove that $x^{3}+y^{3}\leq 2.$ It's stronger: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111000 and still unsolved.

N.T.TUAN wrote: Suppose that positive numbers $x$ and $y$ satisfy $x^{3}+y^{4}\leq x^{2}+y^{3}$. Prove that $x^{3}+y^{3}\leq 2.$ $x^{3}+y^{3}\leq x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{4}-1}{3}=\frac{2}{3}(x^{3}+y^{3})+\frac{2}{3} \implies x^{3}+y^{3}\leq 2.$

sqing wrote: N.T.TUAN wrote: Suppose that positive numbers $x$ and $y$ satisfy $x^{3}+y^{4}\leq x^{2}+y^{3}$. Prove that $x^{3}+y^{3}\leq 2.$ $x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{4}-1}{3}$ How is that true?

sqing wrote: N.T.TUAN wrote: Suppose that positive numbers $x$ and $y$ satisfy $x^{3}+y^{4}\leq x^{2}+y^{3}$. Prove that $x^{3}+y^{3}\leq 2.$ $x^{3}+y^{3}\leq x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{4}-1}{3}=\frac{2}{3}(x^{3}+y^{3})+\frac{2}{3} \implies x^{3}+y^{3}\leq 2.$ $$x^{3}+y^{3}\leq x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{3}-1}{3}=\frac{2}{3}(x^{3}+y^{3})+\frac{2}{3} \implies x^{3}+y^{3}\leq 2.$$

Suppose that positive numbers $x$ and $y$ satisfy $x^2+y^3\leq x+y^2.$ Prove that $$x^3+y^3\leq 2$$Suppose that positive numbers $x$ and $y$ satisfy $x^2+y^4\leq x+y^3.$ Prove that $$x^4+y^4\leq 2$$Suppose that positive numbers $x$ and $y$ satisfy $x^3+y^5\leq x+y^3.$ Prove that $$x^5+y^5\leq 2$$Suppose that positive numbers $x$ and $y$ satisfy $x^3+y^5\leq x^2+y^4.$ Prove that $$x^6+y^6\leq 2$$

sqing wrote: sqing wrote: N.T.TUAN wrote: Suppose that positive numbers $x$ and $y$ satisfy $x^{3}+y^{4}\leq x^{2}+y^{3}$. Prove that $x^{3}+y^{3}\leq 2.$ $x^{3}+y^{3}\leq x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{4}-1}{3}=\frac{2}{3}(x^{3}+y^{3})+\frac{2}{3} \implies x^{3}+y^{3}\leq 2.$ $$x^{3}+y^{3}\leq x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{3}-1}{3}=\frac{2}{3}(x^{3}+y^{3})+\frac{2}{3} \implies x^{3}+y^{3}\leq 2.$$ Sorry could you please explain $x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{3}-1}{3}?$ I just can't seem to grasp my head around that : (

Quantum_fluctuations wrote: sqing wrote: N.T.TUAN wrote: Suppose that positive numbers $x$ and $y$ satisfy $x^{3}+y^{4}\leq x^{2}+y^{3}$. Prove that $x^{3}+y^{3}\leq 2.$ $x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{4}-1}{3}$ How is that true? Brudder wrote: Sorry could you please explain $x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{3}-1}{3}?$ I just can't seem to grasp my head around that : ( $x^{2}+2y^{3}-y^{4}\leqslant \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{3}-1}{3} \Leftrightarrow 0 \leqslant (x-1)^2(2x+1) +(y-1)^2(3y^2+2y+1).$

Brudder wrote: Sorry could you please explain $x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{3}-1}{3}?$ I just can't seem to grasp my head around that : ( By AM-GM.$$x^{3}+y^{3}\leq\frac{2x^{3}+1}{3}=\frac{x^{3}+x^{3}+1}{3}$$$$ 4y^{3}\leq y^{4}+ y^{4}+ y^{4}+1$$ sqing wrote: Suppose that positive numbers $x$ and $y$ satisfy $x^2+y^3\leq x+y^2.$ Prove that $$x^3+y^3\leq 2$$

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Let $a,b,c$ are positive real numsers such that $x^2+y^3+z^4 \geqslant x^3+y^4+z^5.$ (1) Prove that $x^3+y^3+z^3 \leqslant 3.$ (2) Prove that $x^2+y^2+z^2 \leqslant 3.$ https://artofproblemsolving.com/community/c6h3030623p27265297