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This is an old problem of Pham Kim Hung! Prove: $$3\leqq x\sqrt{1+ y^{3}}+ y\sqrt{1+ z^{3}}+ z\sqrt{1+ x^{3}}\leqq 5$$ with $x,\,y,\,z\geqq 0$ & $x+y+z=3$

For the LHS, we have: $$\left \{ \sum\limits_{cyc}x\sqrt{1+ y^{3}} \right \}^{2}= \sum\limits_{cyc}\left \{ x^{2}+ x^{3}z^{2}+ 2\,xy\sqrt{\left ( 1+ y^{3} \right )\left ( 1+ z^{3} \right )} \right \}\geqq \sum\limits_{cyc}\left \{ x^{2}+ 2\,xy \right \}= 9$$

And the RHS, we also have: $$\sum\limits_{cyc}x\sqrt{1+ y^{3}} = \sum\limits_{cyc} x\sqrt{\left ( 1+ y \right )\left ( 1+ y^{2}- y \right )}\leqq \sum\limits_{cyc}\frac{x\left \{ 2+ y^{2} \right \}}{2}= 3+ \frac{1}{2}\sum\limits_{cyc}xy^{2}\leqq 5$$

So, we need to prove: $xy^{2}+ yz^{2}+ zx^{2}\leqq 4$ with $x= 3- y- z$ or $$4- xy^{2}- yz^{2}- zx^{2}= \left \{ 3- y- z \right \}y^{2}- yz^{2}- z\left \{ 3- y- z \right \}^{2}= \left \{ y- 1 \right \}^{2}\left \{ \frac{\underbrace{y+ 4\,z- 5\left ( 3- y- z \right )}_{y+ 4\,z- 5\,x\geqq 0}}{3} \right \}+ \left \{ 3- y- z \right \}\left \{ \frac{\left ( 2\,y+ 2\,z- 3 \right )^{2}+ 3}{4} \right \}\geqq 0$$ with $x= \min\left \{ x,\,y,\,z \right \}$

Any idea? Who can help me with another solution? I hope to see that! Thanks!

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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: $$\sum_{cyc}x\sqrt{1+y^3}=\sum_{cyc}x\sqrt{(1+y)(1-y+y^2)}\leq\frac{1}{2}\sum_{cyc}x(1+y+1-y+y^2)=$$ $$=3+\frac{1}{2}(xy^2+yz^2+zx^2)=3+\frac{1}{2}(xy\cdot y+yz\cdot z+zx\cdot x)\leq$$ $$\leq3+\frac{1}{2}(ab\cdot a+ac\cdot b+bc\cdot c)=3+\frac{1}{2}b(a^2+ac+c^2)\leq$$ $$\leq3+\frac{1}{4}\cdot2b(a+c)^2\leq3+\frac{1}{4}\left(\frac{2b+a+c+a+c}{3}\right)^3=5.$$

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  • $\begingroup$ @Cô_Ngốc_Lớp_Trưởng I solved this problem for you. Why you down-voted? $\endgroup$ – Michael Rozenberg Feb 17 at 6:51

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