How to prove that $1/ ((y+z) x^4) + 1/ ((z+x) y^4) + 1/ ((x+y) z^4) \geq 3/2$ for $x, y, z>0$ such that $xyz=1$? [closed]

Question

How to prove that $\dfrac{1}{(y+z) x^4} + \dfrac{1}{(x+z) y^4} + \dfrac{1}{(y+x) z^4}\geq3/2$ for $x, y, z>0$, such that $xyz=1$?

Answer 1

$\newcommand\tH{\tilde H}$This problem is one of real algebraic geometry, which can be solved purely algorithmically. In Mathematica, such algorithms are implemented by Reduce and similar commands.

Here is a solution with Mathematica:

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Here is a "more human" proof: Let $f(x,y,z)$ stand for the left-hand side of your desired inequality. We want to show that $f(x,y,z)\ge3/2$ if $x,y,z>0$ and $xyz=1$. Equivalently, we want to show that $$h(x,y):=\big(f(x,y,\tfrac1{xy})-\tfrac32\big) 2 x^3 y^3 (x + y) (1 + x^2 y) (1 + x y^2) \\ =2 x^{10} y^{10}+2 x^9 y^8+2 x^8 y^9+2 x^7 y^7-3 x^7 y^6-3 x^6 y^7-3 x^6 y^4+2 x^6 y^2-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^6-3 x^4 y^3+2 x^4 y+2 x^3 y^5-3 x^3 y^4+2 x^2 y^6+2 x y^4+2 y^5\overset{\text{(?)}}\ge0$$ for $x,y>0$.

Consider the partial derivatives $p(x,y):=h_x(x,y)$ and $q(x,y):=h_y(x,y)$. Let $r_1(y)$ and $r_2(x)$ denote the resultants of the polynomials $p(x,y)$ and $q(x,y)$ w.r. to $x$ and $y$, respectively. By symmetry, the real roots of $r_1(y)$ and $r_2(x)$ are the same: $z_1:=0$, $z_2:=1$, and a certain algebraic number $z_3\approx0.8180077783$. Therefore the critical points of $h$ in $(0,\infty)^2$ (if any) are of the form $(z_j,z_k)$ for $j,k\in\{2,3\}$. It is straightforward to check that $h(z_j,z_k)\ge0$ for $j,k\in\{2,3\}$.

So, it remains to check that the boundary values of $h$, near the boundary of the set $[0,\infty]^2$, are $\ge0$.

We have $h(0,y)=2 y^5\ge0$ for $y\ge0$ and $h(x,0)=2x^5\ge0$ for $x\ge0$. This does it for the boundary pieces $\{0\}\times[0,\infty)$ and $[0,\infty)\times\{0\}$.

By symmetry, it remains to show that $\liminf_{x,y\to\infty}h(x,y)\ge0$ and $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. Let $$H_{11}(x,y):=h(\tfrac1x,\tfrac1y)x^{10}y^{10} =2+2 x^{10} y^5+2 x^9 y^6+2 x^8 y^4-3 x^7 y^6+2 x^7 y^5+2 x^6 y^9-3 x^6 y^7-3 x^6 y^4+2 x^5 y^{10}+2 x^5 y^7-6 x^5 y^5+2 x^4 y^8-3 x^4 y^6-3 x^4 y^3-3 x^3 y^4+2 x^3 y^3+2 x^2 y+2 x y^2.$$ Obviously, $H_{11}(0+,0+)=2>0$. So, $\liminf_{x,y\to\infty}h(x,y)\ge0$.

Let $$H_{01}(x,y):=h(\tfrac1x,y)x^{10} =2 x^{10} y^5+2 x^9 y^4+2 x^8 y^6+2 x^7 y^5-3 x^7 y^4-3 x^6 y^6-3 x^6 y^3+2 x^6 y-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^7-3 x^4 y^4+2 x^4 y^2+2 x^3 y^7-3 x^3 y^6+2 x^2 y^9+2 x y^8+2 y^{10}.$$ Removing from the latter expression terms dominated by other terms whenever $x,y\to0$, we end up with having to show that $\liminf_{x,y\to0}\tH(x,y)\ge0$, where $$\tH(x,y):=(2+o(1)) x^5 + (2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 + 2 x^2 y^9 + 2 y^{10},$$ which follows because $$(2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ \ge(2+o(1)) x^5 y^4 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ =x y^8[(2+o(1)) s^4 - (3+o(1)) s^2 + 2 ]>0$$ for all small enough $x,y>0$, where $s:=x/y$. So, $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. $\quad\Box$

Answer 2

Is the "cauchy-schwarz-inequality" tag a guess or a hint? . . .

At any rate, it turns out to be a good start. Let $$ R := \frac1{(y+z) x^4} + \frac1{(z+x) y^4} + \frac1{(x+y) z^4}. $$ We show $xyz = 1 \Rightarrow R > 3/2$, with equality if and only if $(x,y,z) = (1,1,1)$. By Cauchy-Schwarz, $RS \geq T^2$, where $$ S := (y+z) x + (z+x) y + (x+y) z, $$ $$ T := \frac1{x^{3/2}} + \frac1{y^{3/2}} + \frac1{z^{3/2}}. $$ Note that $$ S = 2(yz + zx + xy) = 2\left(\frac1x + \frac1y + \frac1z\right); $$ because $xyz = 1$; in particular, $S \geq 6$ by the AM-GM inequality, with equality $\Leftrightarrow (x,y,z) = (1,1,1)$. By weighted AM-GM, $$ 2 \frac1{x^{3/2}} + 1 \geq \frac3x $$ with equality $\Leftrightarrow x = 1$, and likewise for $y$ and $z$. Therefore $$ 2T \geq \frac32 S - 3, $$ and it remains to prove that $$ S \geq 6 \Rightarrow \frac14 \left(\frac32 S - 3\right)^2 \geq \frac32 S $$ with equality $\Leftrightarrow S = 6$. But this is clear from the factorization $$ \frac14 \left(\frac32 S - 3\right)^2 - \frac32 S = \frac3{16} (S-6) (3S-2) $$ since $S \geq 6 \Rightarrow 3S-2 \geq 16 > 0$. QED