How find the maximum of the value $t$ such $x^3+y^3+z^3+t(xy^2+yz^2+zx^2)\ge (1+t)(x^2y+y^2z+z^2x)$

Question

let $x,y,z\ge 0$, find the maximum of the $t$, such $$x^3+y^3+z^3+t(xy^2+yz^2+zx^2)\ge (1+t)(x^2y+y^2z+z^2x)$$

maybe this take $y=x+u,z=x+t$,But I can't get the answer,because this is very ugly, Thanks

Answer 1

I assumed that it means that we need to find a maximal value of $t$, for which this inequality is true for any non-negatives $x$, $y$ and $z$.

If so, we can say that your way (BW) helps!

Let $x=\min\{x,y,z\}$, $y=x+u$, $z=x+v$ and $u=kv$.

Thus, $u$ and $v$ are non-negatives and $$\sum_{cyc}(x^2+txy^2-(t+1)x^2y)=$$ $$=2(u^2-uv+v^2)x+u^3-(t+1)u^2v+tuv^2+v^3\geq$$ $$\geq u^3-(t+1)u^2v+tuv^2+v^3=v^3(k^3-(t+1)k^2+tk+1).$$ Id est, it's enough to find a maximal value of $t$, for which the inequality $$k^3-(t+1)k^2+tk+1\geq0$$ is true for any non-negative value of $k$ or $$k^3-k^2+1\geq t(k^2-k).$$ Now, it's enough to assume $k>1$ and it's remains to find$$\min_{k>1}\frac{k^3-k^2+1}{k^2-k}.$$ Can you end it now?

I got $$t_{max}=\frac{\sqrt{13+16\sqrt2}-1}{2}\approx2.484...$$

Answer 2

Another way.

I assumed again that it means that we need to find a maximal value of $t$, for which our inequality is true for any non-negatives $x$, $y$ and $z$.

Indeed, our inequality it's $$\sum_{cyc}(x^3+tx^2z-(t+1)x^2y)\geq0$$ or

$$\sum_{cyc}(2x^3+2tx^2z-2(t+1)x^2y)\geq0$$ or $$\sum_{cyc}(2x^3-x^2y-x^2z)\geq(2t+1)\sum_{cyc}(x^2y-x^2z)$$ or $$\sum_{cyc}(x+y)(x-y)^2\geq(2t+1)(x-y)(x-z)(y-z),$$ which says that it's enough to understand, what happens for $x\geq y\geq z$.

But in this case $$(x-y)(x-z)(y-z)\leq(x-y)xy$$ and if we'll decrease our variables by $z$, so the left side will be less and the right side will be greater.

Thus, it's enough to understand, what happens for $z=0$ and we got a previous problem:

To find a maximal value of $t$, for which the inequality $$x^3+y^3+txy^2-(t+1)xy^2\geq0$$ is true for any non-negative numbers $x$ and $y$.