Given three positive numbers $x,y,z$ so that $x+y+z=\frac{3}{2}$. Prove that $28\,x^2y^2z^2+3(x^2y^2+y^2z^2+z^2x^2)\leqq1$ .

Question

Problem. Given three positive numbers $x, y, z$ so that $x+ y+ z= \frac{3}{2}$. Prove that $$28\,x^{2}y^{2}z^{2}+ 3(x^{2}y^{2}+ y^{2}z^{2}+ z^{2}x^{2})\leqq 1$$

Remark. It's the cute twin sister of the following inequality for the same hypothesis given as above: $$\sum\limits_{cyc}\sqrt{\frac{x}{y+ 3\,z}}\geqq x+ y+ z$$ By a.m. - g.m. $$\sum\limits_{cyc}\sqrt{\frac{x}{y+ 3\,z}}\geqq \sum\limits_{cyc}\frac{4\,x(x+ y+ 2\,z)}{4\,x(y+ 3\,z)+ (x+ y+ 2\,z)^{2}}$$ Just need to prove $$\sum\limits_{cyc}\frac{4\,x(x+ y+ 2\,z)}{4\,x(y+ 3\,z)+ (x+ y+ 2\,z)^{2}}\geqq \frac{3}{2}$$

Answer 1

Let $ax+y+z=3u$,$xy+xz+yz=3v^2$ and $xyz=w^3$.

Thus, since $\sum\limits_{cyc}x^2y^2=9v^4-3uw^3,$ we need to prove that $f(w^3)\leq0,$ where $f$ is a convex function.

But the convex function gets a maxinal value for an extreme value of $w^3$,

which happens in the following cases.

1. $w^3\rightarrow0^+$.

Let $z\rightarrow0^+$

Thus, we need to prove that $x^2y^2\leq\frac{1}{3},$ which is true by AM-GM: $$x^2y^2\leq\left(\frac{x+y}{2}\right)^4<\left(\frac{3}{4}\right)^4<\frac{1}{3}.$$ 2. Two variables are equal.

Let $y=x$.

So, $z=\frac{3}{2}-2x,$ where $0<x<\frac{3}{4}$ and we need to prove that $$28x^4\left(\frac{3}{2}-2x\right)^2+3\left(x^4+2x^2\left(\frac{3}{2}-2x\right)^2\right)\leq1$$ or $$(2x-1)^2(2+8x-3x^2+28x^3-56x^4)\geq0,$$ which is true for all $0<x<\frac{3}{4}.$

Indeed, $27x^2-48x^4\geq0$ and $6x^3-8x^4\geq0.$

Thus, it's enough to prove that $$22x^3-30x^2+8x+2\geq0$$ or $$11x^3-15x^2+4x+1\geq0,$$ which is true by AM-GM again: $$11x^3-15x^2+4x+1=5\cdot\frac{11x^3}{5}+3\cdot\frac{4x}{3}+1-15x^2\geq$$ $$\geq9\sqrt[9]{\left(\frac{11x^3}{5}\right)^5\left(\frac{4x}{3}\right)^3}-15x^2=\left(9\sqrt[9]{\left(\frac{11}{5}\right)^5\left(\frac{4}{3}\right)^3}-15\right)x^2>0.$$

Done!