MarkBcc168 wrote: For all non-negative real numbers $x,y,z$ with $x \geq y$, prove the inequality $$\frac{x^3-y^3+z^3+1}{6}\geq (x-y)\sqrt{xyz}.$$ $$x^3-y^3+z^3+1=(x-y)^3+3xy(x-y)+z^3+1\geq 6\sqrt[6]{ (x-y)^3x^3y^3(x-y)^3z^3}=6\sqrt{ (x-y)xyz}.$$

MarkBcc168 wrote: For all non-negative real numbers $x,y,z$ with $x \geq y$, prove the inequality $$\frac{x^3-y^3+z^3+1}{6}\geq (x-y)\sqrt{xyz}.$$ Equality holds when $x=\frac{\sqrt{5}+1}{2}, y=\frac{\sqrt{5}-1}{2} , z=1.$ Nice problem. Proof of Zhangyanzong:

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Is this correct ? : Since \[ (x-y)\sqrt{xyz} \leq (x-y)(\dfrac{xy+z}{2}) \]It suffices to show that $$x^3-y^3+z^3 \geq 3(x-y)(xy+z) $$. Considering this as a function of $z$ , using derivatives, the minima occurs for $z=\sqrt{x-y}$. If we put $t^2=x-y$ then the inequality is equivalent to $$t^6-2t+1 \geq 0$$, which is true.$\blacksquare$ Edit: Actually the last inequality is not true for $\approx 0.5 \leq t <1$.But anyways it proves a stronger inequality for many $t$ and a counter example for others...

hellomath010118 wrote: Edit: Actually the last inequality is not true for $\approx 0.5 \leq t <1$.But anyways it proves a stronger inequality for many $t$ and a counter example for others... Nope. Of course it does not provide a counterexample to the original problem but to your simplified version. So our problem is $A$ and your problem is $B$. You showed that $B \Rightarrow A$ but then you show that $B$ is wrong. Of course this does not say anything about $A$... (An easy counterexample to your problem $B$ which is the inequality $x^3-y^3+z^3 \ge 3(x-y)(xy+z)$ would be $x=1, y=\frac{1}{2}, z=\frac{3}{4}$).

Tintarn wrote: hellomath010118 wrote: Edit: Actually the last inequality is not true for $\approx 0.5 \leq t <1$.But anyways it proves a stronger inequality for many $t$ and a counter example for others... Nope. Of course it does not provide a counterexample to the original problem but to your simplified version. So our problem is $A$ and your problem is $B$. You showed that $B \Rightarrow A$ but then you show that $B$ is wrong. Of course this does not say anything about $A$... (An easy counterexample to your problem $B$ which is the inequality $x^3-y^3+z^3 \ge 3(x-y)(xy+z)$ would be $x=1, y=\frac{1}{2}, z=\frac{3}{4}$). I meant counterexample to the stronger inequality not original one AND you can verify this example that you provide was already declared by me if you see clearly.

hellomath010118 wrote: ...AND you can verify this example that you provide was already declared by me if you see clearly. Sure, this is of course how I found it. I just wanted to make clear that this was neither a proof nor a disproof of the original problem since you seemed to ask for it by asking "Is it correct?".

For all non-negative real numbers $x,y,z$ with $x \geq y$, prove the inequality $$\frac{x^3-y^3+z^3+1}{6}\geq\frac{1}{2}(x-y)(xy+z)\geq (x-y)\sqrt{xyz}.$$Equality holds when $x=\frac{\sqrt{5}+1}{2}, y=\frac{\sqrt{5}-1}{2} , z=1.$ See＃4

Mr. Sqing I think that the term in the middle should not be there as a counterexample has been generated by #8.

Classic example of working backwards: We have $$(x-y)^3+z^3+1 \geq (x-y)^3+z^{\frac{3}{2}}+z^{\frac{3}{2}} \geq 3(x-y)z \implies \frac{x^3-y^3+z^3+1}{6} \geq \frac{(x-y)z+x^2y-y^2x}{2} \geq \sqrt{(x-y)^2(xyz)}$$

can someone give me advice on how to think on such problems?

va2010 made a great pdf for inequalities. One of the ideas in the pdf is to form a simpler inequality, and backtrack to prove the actual inequality (you'll understand once you do the problems). This was the idea I used.

Plops wrote: va2010 made a great pdf for inequalities. One of the ideas in the pdf is to form a simpler inequality, and backtrack to prove the actual inequality (you'll understand once you do the problems). This was the idea I used. thanks a lot

MarkBcc168 wrote: For all non-negative real numbers $x,y,z$ with $x \geq y$, prove the inequality $$\frac{x^3-y^3+z^3+1}{6}\geq (x-y)\sqrt{xyz}.$$ we assume $x=y+a$ which $a$ is non-negetive, then $a^3+3ay(a+y)+z^3+1\geq 6a\sqrt{y(y+a)z}$ done.

E.A.K wrote: MarkBcc168 wrote: For all non-negative real numbers $x,y,z$ with $x \geq y$, prove the inequality $$\frac{x^3-y^3+z^3+1}{6}\geq (x-y)\sqrt{xyz}.$$ we assume $x=y+a$ which $a$ is non-negetive, then $a^3+3ay(a+y)+z^3+1\geq 6a\sqrt{y(y+a)z}$ done. why is it true?

Note that we want to prove $x^3-y^3+z^3+1 \ge 6\sqrt[6]{(x-y)^6x^3y^3z^3}$. $x^3-y^3+z^3+1 = (x-y)^3 + 3xy(x-y) + z^3 + 1 = (x-y)^3 + xy(x-y) + xy(x-y) + xy(x-y) + z^3 + 1 \ge 6\sqrt[6]{(x-y)^6x^3y^3z^3}$.