Ferid.---. wrote: The real numbers $x,y,z$ satisfy $x^2+y^2+z^2=3.$ Prove that the inequality $x^3-(y^2+yz+z^2)x+yz(y+z)\le 3\sqrt{3}.$ and find all triples $(x,y,z)$ for which equality holds. Note that $x^3-(y^2+yz+z^2)x+yz(y+z)=(x+y+z)(x-y)(x-z)$ So we will consider just the case that exactly two or none of {x+y+z, x-y, x-z} are negative. Case1 : If none of them are negative then by AM-GM, $(x+y+z)(x-y)(x-z)\leq (\frac{(x+y+z)+(x-y)+(x-z)}{3})^3=x^3$ where $|x|\leq \sqrt{3}, x^3\leq |x|^3\leq 3\sqrt{3}$, equality holds when $(x,y,z)=(\sqrt{3},0,0)$ Case2 : If exactly two of them are negative, $(x+y+z,x-y,x-z)=(+,-,-),(-,+,-),(-,-,+)$ For $(-,+,-)$ : $(x+y+z)(x-y)(x-z)\leq (\frac{(-x-y-z)+(x-y)+(z-x)}{3})^3=(\frac{-x-2y}{3})^3$ but $\frac{-x-2y}{3}\leq max{|x|,|y|,|z|}\leq \sqrt{3}$ so we are done. For $(-,-,+)$ : Likewise, we can do then same method as the upper lines. For $(+,-,-)$ : $(x+y+z)(x-y)(x-z)\leq (\frac{(x+y+z)+(y-x)+(z-x)}{3})^3=(\frac{2y+2z-x}{3})^3$ so its suffice to prove that $2y+2z-x\leq 3\sqrt{3}$ we can see that $(y+z)^2\leq 2(y^2+z^2)=2(3-x^2), y+z\leq \sqrt{2(3-x^2)}$ it remains to prove $2\sqrt{2(3-x^2)}\leq x+3\sqrt{3}\iff 8(3-x^2)\leq x^2+6\sqrt{3}x+27\iff 0\leq 3(\sqrt{3}x+1)^2$ whch is true Done

Ferid.---. wrote: The real numbers $x,y,z$ satisfy $x^2+y^2+z^2=3.$ Prove that the inequality $x^3-(y^2+yz+z^2)x+yz(y+z)\le 3\sqrt{3}.$ and find all triples $(x,y,z)$ for which equality holds.

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The non-negative real numbers $x,y,z$ satisfy $x+y+z=3.$ Find the maximum value of $x^3-(y^2+yz+z^2)x+yz(y+z).$

NahTan123xyz wrote: Ferid.---. wrote: The real numbers $x,y,z$ satisfy $x^2+y^2+z^2=3.$ Prove that the inequality $x^3-(y^2+yz+z^2)x+yz(y+z)\le 3\sqrt{3}.$ and find all triples $(x,y,z)$ for which equality holds. Note that $x^3-(y^2+yz+z^2)x+yz(y+z)=(x+y+z)(x-y)(x-z)$ So we will consider just the case that exactly two or none of {x+y+z, x-y, x-z} are negative. Case1 : If none of them are negative then by AM-GM, $(x+y+z)(x-y)(x-z)\leq (\frac{(x+y+z)+(x-y)+(x-z)}{3})^3=x^3$ where $|x|\leq \sqrt{3}, x^3\leq |x|^3\leq 3\sqrt{3}$, equality holds when $(x,y,z)=(\sqrt{3},0,0)$ Case2 : If exactly two of them are negative, $(x+y+z,x-y,x-z)=(+,-,-),(-,+,-),(-,-,+)$ For $(-,+,-)$ : $(x+y+z)(x-y)(x-z)\leq (\frac{(-x-y-z)+(x-y)+(z-x)}{3})^3=(\frac{-x-2y}{3})^3$ but $\frac{-x-2y}{3}\leq max{|x|,|y|,|z|}\leq \sqrt{3}$ so we are done. For $(-,-,+)$ : Likewise, we can do then same method as the upper lines. For $(+,-,-)$ : $(x+y+z)(x-y)(x-z)\leq (\frac{(x+y+z)+(y-x)+(z-x)}{3})^3=(\frac{2y+2z-x}{3})^3$ so its suffice to prove that $2y+2z-x\leq 3\sqrt{3}$ we can see that $(y+z)^2\leq 2(y^2+z^2)=2(3-x^2), y+z\leq \sqrt{2(3-x^2)}$ it remains to prove $2\sqrt{2(3-x^2)}\leq x+3\sqrt{3}\iff 8(3-x^2)\leq x^2+6\sqrt{3}x+27\iff 0\leq 3(\sqrt{3}x+1)^2$ whch is true Done That's exactly how I solved it, but I didn't have time to write all the equality cases :/ . How many points do I lose for that?

Published here too : https://www.cut-the-knot.org/m/Algebra/EuroMathCup2017JuniorProblem4.shtml

Can this problem be solved with Langrange Multipliers?

Yes it can be, but it won't be any pleasure.

sqing wrote: The non-negative real numbers $x,y,z$ satisfy $x+y+z=3.$ Find the maximum value of $x^3-(y^2+yz+z^2)x+yz(y+z).$ $$(x+y+z)(y^2+z^2+yz+3xy+3xy)\ge 0 \Rightarrow$$$$ x^3-(y^2+yz+z^2)x+yz(y+z) \le (x+y+z)^3=27$$

Ferid.---. wrote: The real numbers $x,y,z$ satisfy $x^2+y^2+z^2=3.$ Prove that the inequality $x^3-(y^2+yz+z^2)x+yz(y+z)\le 3\sqrt{3}.$ and find all triples $(x,y,z)$ for which equality holds. $$3(x^2+y^2+z^2)=(x+y+z)^2+(x-y)^2+(y-z)^2+(z-x)^2\ge $$$$ \ge (x+y+z)^2+(x-y)^2+(x-z)^2 \ge 3 \sqrt[3] {(x+y+z)^2(x-y)^2(x-z)^2} \Rightarrow$$$$ (x+y+z)^2(x-y)^2(x-z)^2 \le 27$$

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