Answer is 9. Am I right?

let$ x=1,y=\frac{1}{2}$ and$ z=0$ . then $ C \leq \frac{5}{2}$ ineq becomes : $ 2(x^3+y^3+z^3) + 5(xy^2+yz^2+zx^2) \geq 7(x^2y+y^2z+z^2x)$

Applying CID (Cyclic inequality oF degree 3) theorem, we can let $ c = 0$ in the inequality. It becomes \[ x^3 + y^3 + cx^2y\ge (c + 1) xy^2. \] Thus, we have to find the minimal value of \[ f(y) = \frac {y^3 - y^2 + 1}{y^2 - y} = y + \frac {1}{y(y - 1)} \] when $ y > 1$. It is easy to find that \[ f'(y) = 0 \Leftrightarrow 2y - 1 = (y(y - 1))^2 \Leftrightarrow y^4 - 2y^3 + y^2 - 2y + 1 = 0. \] Solving this symmetric equation gives us: \[ y + \frac {1}{y} = 1 + \sqrt {2} \Rightarrow y = \frac{1 + \sqrt {2} + \sqrt {2\sqrt {2} - 1}}{2} \] Thus we found the best value of $ C$ is \[ y + \frac {1}{y(y - 1)} = \frac {1 + \sqrt {2} + \sqrt {2\sqrt {2} - 1}}{2} + \frac {1}{\sqrt {\sqrt {2} + \sqrt {2\sqrt {2} - 1}}}\approx 2.4844 \]

hungkhtn wrote: Applying CID (Cyclic inequality oF degree 3) theorem, we can let $ c = 0$ in the inequality. You mean $ z = 0$? Can you give a statement and proof?

t0rajir0u wrote: hungkhtn wrote: Applying CID (Cyclic inequality oF degree 3) theorem, we can let $ c = 0$ in the inequality. You mean $ z = 0$? Can you give a statement and proof? Here you are: http://www.mathlinks.ro/viewtopic.php?p=1130901#1130901

A very strange computational problem to give in an olympiad... anyways to see why we only need to check when $ z=0$ is a simple argument. note that the inequality can be rewritten as $ \sum (x+y)(x-y)^2\geq (2c+1)(x-y)(y-z)(x-z)$ so decreasing each number by $ k$ would only decrease the RHS and leave LHS fixed.

Albanian Eagle wrote: A very strange computational problem to give in an olympiad... anyways to see why we only need to check when $ z = 0$ is a simple argument. note that the inequality can be rewritten as $ \sum (x + y)(x - y)^2\geq (2c + 1)(x - y)(y - z)(x - z)$ so decreasing each number by $ k$ would only decrease the RHS and leave LHS fixed. Yes, that is simpler, although the idea is exactly the same as the classical entirely-mixing-variable method.

hungkhtn wrote: $ \frac {1 + \sqrt {2} + \sqrt {2\sqrt {2} - 1}}{2} + \frac {1}{\sqrt {\sqrt {2} + \sqrt {2\sqrt {2} - 1}}}\approx 2.4844$ This answer is correct. Just in case, it can be rewritten as $ \frac{1}{2}\left(\sqrt {13+16\,\sqrt {2}}-1\right)$. It's the root of the polynomial $ k^4+2k^3-5k^2-6k-23$. Sergey Markelov

hi markelov how did you get the equation $ k^4+2k^3-5k^2-6k-23=0$

I found nice solution with Vornicu Schur Inequality \[ C = - \frac12 +\frac {7}{2\sqrt{28\sqrt {2} - 35}} - \frac{1}{ -\frac12 + \frac {1}{14}\sqrt {28\sqrt {2} - 35}} -\frac {1}{\frac12 +\frac {1}{14}\sqrt {28\sqrt{2} -35}}\approx 2.48443533 \]

$ x^3+y^3+z^3+\frac{\sqrt{13+16\sqrt{2}}-1}{2}\left(xy^2+yz^2+zx^2\right)$ $ -\frac{\sqrt{13+16\sqrt{2}}+1}{2}\left(x^2y+y^2z+z^2x\right)$ $ =\left(\frac{1+\sqrt{8\sqrt{2}-11}}{2}y+\frac{1-\sqrt{8\sqrt{2}-11}}{2}z-x\right)\left(x+\frac{\sqrt{5+4\sqrt{2}}-1}{2}y-\frac{\sqrt{5+4\sqrt{2}}+1}{2}z\right)^2$ $ +2x\left(x^2+y^2+z^2-yz-zx-xy\right)\geq0,$ which is clearly true for $ x = \min\{x,y,z\}.$

Litlle 1000t wrote: Find the maximum number $ C$ such that for any nonnegative $ x,y,z$ the inequality $ x^3 + y^3 + z^3 + C(xy^2 + yz^2 + zx^2) \ge (C + 1)(x^2 y + y^2 z + z^2 x)$ holds. Prove that for all nonnegative numbers $a,b,c$ the following inequality holds $$ x^3+y^3+z^3+\frac{\sqrt{13+16\sqrt{2}}-1}{2}\left(xy^2+yz^2+zx^2\right)\geq\frac{\sqrt{13+16\sqrt{2}}+1}{2}\left(x^2y+y^2z+z^2x\right).$$Proof of dragonheart6:

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