(Pham Kim Hung). Given three non-negative numbers $x, y, z$ so that $x+ y+ z= 3$. Prove that: $$3\leqq x\sqrt{1+ y^{3}}+ y\sqrt{1+ z^{3}}+ z\sqrt{1+ x^{3}}\leqq 5$$
For the left inequality, we have: $$\left ( \sum\limits_{cyc} x\sqrt{1+ y^{3}} \right )^{2}= \sum\limits_{cyc}\left (\!x^{2}+ x^{3}z^{2}+ 2xy\sqrt{(1+ y^{3})(1+ z^{3})}\!\right )\geqq \sum\limits_{cyc}(x^{2}+ 2xy)= 9$$ For the left inequality, we have: $$\sum\limits_{cyc} x\sqrt{1+ y^{3}}= \sum\limits_{cyc} x\sqrt{(1+ y)(1+ y^{2}- y)}\leqq \sum\limits_{cyc}\frac{x(2+ y^{2})}{2}= 3+ \frac{1}{2}\sum\limits_{cyc}xy^{2}\leqq 5$$ We will prove $xy^{2}+ yz^{2}+ zx^{2}\leqq 4$ so that $x= 3- y- z$ and $x\equiv \min\{x, y, z\}$, equivalent to $$4- \sum\limits_{cyc}xy^{2}= (\!y- 1\!)^{2}\left (\!\frac{\overbrace{y+ 4z- 5(\!3- y- z\!)}^{y+ 4z- 5x\geqq 0}}{3}\!\right )+ (\!3- y- z\!)\left (\!\frac{(\!2y+ 2z- 3\!)^{2}+ 3}{4}\!\right )\geqq 0$$ Anyone's idea? Who can help me another solution?
I think you mean $x+y+z=3.$
The left inequality: $$\sum_{cyc}x\sqrt{1+y^3}\geq\sum_{cyc}x=3.$$ The right inequality.
Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c.$
Thus, by AM-GM, Rearrangement and AM-GM we obtain: $$\sum_{cyc}x\sqrt{1+y^3}=\sum_{cyc}x\sqrt{(1+y)(1-y+y^2)}\leq\frac{1}{2}\sum_{cyc}x(1+y+1-y+y^2)=$$ $$=3+\frac{1}{2}(xy^2+yz^2+zx^2)=3+\frac{1}{2}(xy\cdot y+yz\cdot z+zx\cdot x)\leq$$ $$\leq3+\frac{1}{2}(ab\cdot a+ac\cdot b+bc\cdot c)=3+\frac{1}{2}b(a^2+ac+c^2)\leq$$ $$\leq3+\frac{1}{4}\cdot2b(a+c)^2\leq3+\frac{1}{4}\left(\frac{2b+a+c+a+c}{3}\right)^3=5.$$