Problematic inequality & hint

Question

I would like to ask for hint for proving following inequality: $$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$ for all $x>0$, $y>0$, $z>0$ such that $xyz=1$. Generally, I tried to find some elementary solution, but even with some calculus I didn't solve it.

Edit. My attempt: Let $f:(0,\infty)\times(0,\infty)\rightarrow \mathbb{R}$ be function defined by equality: $$ f(x,y)=x^3(1+x)+y^3(1+y)+\frac{1}{x^3y^3}(1+\frac{1}{xy})-\frac{3}{4}(1+x)(1+y)(1+\frac{1}{xy})$$ and I calculated partial derivatives: $$\frac{\partial f}{\partial x }(x,y)=-\frac{4}{x^5y^4}-\frac{3}{x^4y^3}+4x^3+\frac{3(y+1)}{4x^2y}+3x^2-\frac{3(y+1)}{4}$$ $$\frac{\partial f}{\partial y }(x,y)=-\frac{4}{y^5x^4}-\frac{3}{y^4x^3}+4y^3+\frac{3(x+1)}{4y^2x}+3y^2-\frac{3(x+1)}{4}$$ I noticed that: $$\frac{\partial f}{\partial x }(1,1)=\frac{\partial f}{\partial x }(1,1)=0.$$ I began to wonder if the function is convex (i.e. by demonstrating that $g(w)=w^3(1+w)$ is convex for positive values and $h(x,y)=(1+x)(1+y)(1+\frac{1}{xy})$ is also concave for positive x and y, with latter I had some calculation trouble).

Answer 1

Since $x^3$ is convex for $x\ge 0$, we have that $x^3 \ge 1 + 3(x-1)$, using the tangent at $x=1$.(*) With this observation, it is enough to prove that $$ \sum_{cyc} (3x-2)(1+x)\geq \frac{3}{4}(1+x)(1+y)(1+z) $$ Expanding the terms, and using $x y z = 1$, gives the equivalent $$ -30 + \sum_{cyc} x + 12 \sum_{cyc} x^2 - 3 \sum_{cyc} xy \ge 0 $$ Now note $$ 3 \sum_{cyc} x^2 - 3 \sum_{cyc} xy = 3(\frac12 \sum_{cyc} x^2 +\frac12 \sum_{cyc} y^2 - \sum_{cyc} xy ) = \frac{3}{2}(\sum_{cyc} (x-y)^2) \ge 0 $$ So it is enough to show $$ -30 + \sum_{cyc} x + 9 \sum_{cyc} x^2 \ge 0 $$ For the two sums in here, we have by AM-GM: $\sum_{cyc} x \ge 3 \sqrt[3]{xyz} = 3$ and by Titu's Lemma (Cauchy - Schwarz) $\sum_{cyc} x^2 \ge \frac13 (\sum_{cyc} x )^2 \ge \frac93= 3$

So we have $$ -30 + \sum_{cyc} x + 9 \sum_{cyc} x^2 \ge -30 + 3 +9 \cdot 3 = 0 $$ which proves the claim. $\qquad \Box$

(*) Alternatively, if you do not want calculus arguments, $x^3 \ge 1 + 3(x-1)$ can be shown elementary. Let $y = x-1$. Then we have $x^3 = (1+y)^3 = 1 + 3y + y^2(3+y)$ and this is $\ge 1 + 3y$ as long as $y \ge -3$ or $x \ge -2$, which is given since we are interested in $x \ge 0$.

Answer 2

We need to prove that: $$\sum_{cyc}(4x^4+4x^3)\geq\sum_{cyc}(2+3x+3xy),$$ which is true by Muirhead.

Indeed, we need to prove that: $$\sum_{cyc}(3x^4+3x^3+x^4+x^3)\geq\sum_{cyc}(3xy+3x+1+1).$$ Now, $$\sum_{cyc}x^4\geq\sum_{cyc}xy$$ it's $$\sum_{cyc}x^4\geq\sum_{cyc}xy\sqrt[3]{x^2y^2z^2}$$ or $$\sum_{cyc}x^4\geq\sum_{cyc}x^{\frac{5}{3}}y^{\frac{5}{3}}z^{\frac{2}{3}},$$ which is true by Muirhead because $$\left(4,0,0\right)\succ\left(\frac{5}{3},\frac{5}{3},\frac{2}{3}\right).$$ Also, $$\sum_{cyc}x^3\geq\sum_{cyc}x$$ it's $$\sum_{cyc}x^3\geq\sum_{cyc}x^{\frac{5}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}},$$ which is true by Muirhead again because $$(3,0,0)\succ\left(\frac{5}{3},\frac{2}{3},\frac{2}{3}\right).$$ Also, by AM-GM $$\sum_{cyc}x^4\geq3\sqrt[3]{x^4y^4z^4}=3=\sum_{cyc}1$$ and $$\sum_{cyc}x^3\geq3xyz=3=\sum_{cyc}1.$$

Answer 3

(Edited version). By AM-GM we have:

$$ \sum_{cyc}(\frac{x^3}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8})$$ $$ \geq \sum_{cyc}3 \sqrt[3]{\frac{x^3}{(1+y)(1+z)} \cdot \frac{1+y}{8} \cdot \frac{1+z}{8}}$$ $$=\frac{3}{4}(x+y+z)$$ we have: $$ \sum_{cyc}(\frac{x^3}{(1+y)(1+z)}\geq \frac{3}{4}(x+y+z) - \sum_{cyc}(\frac{1+y}{8} + \frac{1+z}{8}) $$ $$ =\frac{2}{4}(x+y+z)-\frac{3}{4}\geq \frac{1}{2}\cdot 3 \sqrt[3]{xyz}-\frac{3}{4}=\frac{3}{4}. $$Q.E.D.