Hi, this is my solution: https://ibb.co/CmJkPd5 Sorry I cannot put LaTeX here but this link should be fine.

LaSeineFirenze wrote: Hi, this is my solution: see here Sorry I cannot put LaTeX here but this link should be fine. Interesting

jasperE3 wrote: Show that if real numbers $x<1<y$ satisfy the inequality $$2\log x+\log(1-x)\ge3\log y+\log(y-1),$$then $$x^3+y^3<2$$ $$2\log x+\log(1-x)\ge3\log y+\log(y-1)\iff x^2 +y^3\geq x^3 +y^4$$ Russian 1999: Let $x$ and $y$ are positive real numbers such that $x^2 +y^3\geq x^3 +y^4.$ Prove that$$x^3 + y^3\leq 2$$Hungary-Israel Binational 2002 sqing wrote: $x^{3}+y^{3}\leq x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{4}-1}{3}=\frac{2}{3}(x^{3}+y^{3})+\frac{2}{3} \implies x^{3}+y^{3}\leq 2.$