Proving Inequality with Positive Real Numbers $x,\,y,\,z$

[anemone]

Let $x,\,y,\,z$ be positive real numbers such that $xy+yz+zx=3$.

Prove the inequality $(x^3-x+5)(y^5-y^3+5)(z^7-z^5+5)\ge 125$.

 

[anemone]

Here is the solution that I found online and would like to share with MHB:

Spoiler

For any real numbers $a$, the numbers $a-1,\,a^2-1,\,a^3-1,\,a^5-1$ are of the same sign.

Therefore

$(a-1)(a^2-1)\ge 0$, $(a^2-1)(a^3-1)\ge 0$ and $(a^2-1)(a^5-1)\ge 0$.

i.e.

$x^3-x^2-x+1\ge 0$

$y^5-y^3-y^2+1\ge 0$

$z^7-z^5-z^2+1\ge 0$

So it follows that

$x^3-x+5\ge x^2+4$, $y^5-y^3+5\ge y^2+4$ and $z^7-z^5+5\ge z^2+4$

Multiplying these inequalities gives

$(x^3-x+5)(y^5-y^3+5)(z^7-z^5+5)\ge(x^2+4)(y^2+4)(z^2+4) \tag{1}$

We will prove that

$(x^2+4)(y^2+4)(z^2+4)\ge 25(xy+yz+xz+2) \tag{2}$

We have

$\begin{align*}(x^2+4)(y^2+4)(z^2+4)&=x^2y^2z^2+4(x^2y^2+y^2z^2+z^2x^2)+16(x^2+y^2+z^2)+64\\&=x^2y^2z^2+(x^2+y^2+z^2)+2+4(x^2y^2+y^2z^2+z^2x^2+3)+15(x^2+y^2+z^2)+50---(3)\end{align*}$

By the inequalities

$(x-y)^2+(y-z)^2+(z-x)^2\ge 0$ and

$(xy-1)^2+(yz-1)^2+(zx-1)^2\ge 0$ we obtain

$x^2+y^2+z^2\ge xy+yz+zx\tag{4}$ and

$x^2y^2+y^2z^2+z^2x^2+3\ge 2(xy+yz+zx)\tag{5}$

We will prove that

$x^2y^2z^2+(x^2+y^2+z^2)+2\ge 2(xy+yz+zx)\tag{6}$

Note that if we have $a,\,b,\,c>0$ then $3abc+a^3+b^3+c^3\ge 2((ab)^{\dfrac{3}{2}}+(bc)^{\dfrac{3}{2}}+(ca)^{\dfrac{3}{2}})$.

Its proof followed by Schur's inequality and AM-GM inequality.

For $a=x^{\dfrac{2}{3}}$, $b=y^{\dfrac{2}{3}}$ and $c=z^{\dfrac{2}{3}}$, we deduce

$3(xyz)^{\dfrac{2}{3}}+x^2+y^2+z^2\ge 2(xy+yz+zx)$

Therefore, it suffices to prove that

$x^2y^2z^2+2\ge 3(xyz)^{\dfrac{2}{3}}$, which follows immediately by $AM>GM$.

Thus we have proved inequality (6).

Now, by (3), (4), (5) and (6) we obtain inequality (2).

Finally by (1) and (2) and since $xy+yz+zx=3$ we obtain the required inequality.

Equality occurs iff $x=y=z=1$.