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If $x,y,z>0$ and $\sum_{cyc}^{} x^2 (1-x) \ge 0 $, I have to prove that:

$\sum_{cyc}^{} x^2 y^2 (1-xy) \ge 0 $

Then, if $\sum_{cyc}^{} x=3,$ then I have to prove

$\sqrt [8] \frac {\sum_{cyc}^{} x^3}{3} \le \frac {3}{\sum_{cyc}^{} xy}$

I tried to use Jensen's inequality, for $f(x)=x^2 (1-x),$ $g(x)=x^3$, $f_1(x,y)=x^2 y^2 (1-xy)$, $g_1(x,y)=\frac{1}{xy}$ without any success. I tried also C-S inequality but I was wondering if $1-x, 1-xy>0$. Any help? Thank you

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Your second inequality.

We need to prove that $$\left(\frac{x+y+z}{3}\right)^{19}\geq\left(\frac{xy+xz+yz}{3}\right)^8\frac{x^3+y^3+z^3}{3}.$$ We'll prove a stronger inequality:

Let $x$, $y$ and $z$ be positives. Prove that $$\left(\frac{x+y+z}{3}\right)^{16}\geq\left(\frac{xy+xz+yz}{3}\right)^{6.5}\frac{x^3+y^3+z^3}{3}.$$ Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$ and $xyz=w^3$.

Thus, we need to prove that $$u^{16}\geq v^{13}(9u^3-9uv^2+w^3),$$ which says that it's enough to prove the last inequality for a maximal value of $w^3$,

which happens for equality case of two variables.

Since the last inequality is homogeneous, we can assume $y=z=1$ and we need to prove that $$\left(\frac{x+2}{3}\right)^{16}\geq\left(\frac{2x+1}{3}\right)^{6.5}\frac{x^3+2}{3}$$ or $f(x)\geq0,$ where $$f(x)=16\ln(x+2)-6.5\ln(2x+1)-\ln(x^3+2)-8.5\ln3.$$ Now, $$f'(x)=\frac{(x-1)(13x^3-12x^2-18x+20)}{(x+2)(2x+1)(x^3+2)},$$ which gives $x_{min}=1$ and we are done!

Your first inequality:

The condition gives $$1\geq\frac{x^3+y^3+z^3}{x^2+y^2+z^2}.$$ Thus, we need to prove that $$\sum_{cyc}x^2y^2\geq\sum_{cyc}x^3y^3,$$ for which it's enough to prove that $$\frac{(x^2y^2+x^2z^2+y^2z^2)(x^3+y^3+z^3)^2}{(x^2+y^2+z^2)^2}\geq x^3y^3+x^3z^3+y^3z^3$$ or $$\sum_{cyc}(x^8y^2+x^8z^2-x^7y^3-x^7z^3+x^6y^2z^2-x^4y^3z^3)\geq0,$$ which is true by Muirhead.

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