$S_p = \dfrac{1}{(y+z) x^p} + \dfrac{1}{(z+x) y^p} + \dfrac{1}{(x+y) z^p} \geq \dfrac{3}{2}$

Question

Main Problem: Let $x,y,z$ be positive real numbers and $xyz = 1$. For any $p$ real number, let's define $$ S_p = \dfrac{1}{(y+z) x^p} + \dfrac{1}{(z+x) y^p} + \dfrac{1}{(x+y) z^p} $$

Prove (or disprove) that for all $p\geq 2$ values, $S_p\geq \dfrac{3}{2}$.

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My Efforts:

It would be helpful if I explained to the readers how I created the problem and what methods I tried.

• We show that $S_4\geq \dfrac{3}{2}$ in MSE here.

• Also $S_3\geq \dfrac{3}{2}$ is a problem from IMO-1995. (There are many different solution method in the link.)

• For $p=2$, I proved $S_2 = \dfrac{1}{(y+z) x^2} + \dfrac{1}{(z+x) y^2} + \dfrac{1}{(x+y) z^2} \geq \dfrac{3}{2}$.

Proof for $p=2$ Case: For $a,b,c$ real numbers $a^2 + b^2 + c^2 \geq ab + bc + ca$. Also $xyz=1$ is given and so we can write $(xy)^2 + (yz)^2 + (zx)^2 \geq xy\cdot yz + zx\cdot xy + yz\cdot zx = x+y+z$. By Sedrakyan's Form of Cauchy-Schwarz inequality and by the above inequality, we get

$ S_2 = \dfrac{1}{(y+z) x^2} + \dfrac{1}{(z+x) y^2} + \dfrac{1}{(x+y) z^2} \\ = \dfrac{(xyz)^2}{(y+z) x^2} + \dfrac{(xyz)^2}{(z+x) y^2} + \dfrac{(xyz)^2}{(x+y)z^2} = \dfrac{(yz)^2}{(y+z)} + \dfrac{(zx)^2}{(z+x)} + \dfrac{(xy)^2}{(x+y)} \\ \geq \dfrac{\left(xy + yz + zx \right)^2}{2(x+y+z)} = \dfrac{(xy)^2 + (yz)^2 + (zx)^2 + 2(x+y+z)}{2(x+y+z)} \\ \geq \dfrac{(x+y+z)+2(x+y+z)}{2(x+y+z)}=\dfrac{3}{2}.$

$p=1$ Case: If we take $x=y=2$, $z=\dfrac{1}{4}$, we can calculate the value $S_1 = \dfrac{13}{9} < \dfrac{3}{2}$. In this case, inequality is not achieved.

$p=\dfrac{3}{2}$ Case: If we take $x=y=4$, $z=\dfrac{1}{16}$, we can calculate the value $S_{3/2} = \dfrac{4}{65} + 8 > \dfrac{3}{2}$. In this case, inequality seems achieved. So, I felt that the inequality should be true when $p>1$. This is how the idea of the Main Problem came about. On the other hand, by Reza Rajaei's comment $(x,y,z) = (50,\dfrac{1}{50},1)$ and for $p=1.1$, we see that $S_{1.1}<\dfrac{3}{2}$.

• WLOG, $0 <x \leq y \leq z$. Then $x^p \leq y^p \leq z^p$ and $x+y\leq x+z \leq y+z$. I thought of using Chebyshev's sum inequality.

$$ S_p = \dfrac{1}{(y+z) x^p} + \dfrac{1}{(z+x) y^p} + \dfrac{1}{(x+y) z^p} \leq \dfrac{1}{3}\cdot \left( \dfrac{1}{x^p} + \dfrac{1}{y^p} + \dfrac{1}{z^p} \right) \left( \dfrac{1}{y+z} + \dfrac{1}{z+x} + \dfrac{1}{x+y} \right) .$$ This equality is correct, but unuseful unfortunalely. Because, the inequality is in the opposite direction.

• I tried homogenizing $S_p$. $$ S_p = \dfrac{(xyz)^{(p+1)/3}}{(y+z) x^p} + \dfrac{(xyz)^{(p+1)/3}}{(z+x) y^p} + \dfrac{(xyz)^{(p+1)/3}}{(x+y) z^p} $$

and now it is homogenize. I had some more futile efforts and then I got stuck. Thank you for your interest.

Answer 1

Oh this one is not that bad. First, substitute $x\to\frac 1x$, etc for simplicity to obtain an equivalent inequality: $$\sum\dfrac{x^{p-1}}{y+z}\geq \frac 32.$$ Then, do the natural Cauchy-Schwarz to obtain: $$\sum\dfrac{x^{p-1}}{y+z}\sum (y+z)\geq \left(\sum x^{\tfrac{p-1}{2}}\right)^2.$$ But if $p\geq 2$, then $\dfrac{p-1}{2}\geq 1$ and it's an easy exercise to prove: $$x^m+y^m+z^m\geq (xyz)^{\frac{m-1}{3}}(x+y+z)$$ when $m\geq 1.$ Combine all these together and your inequality follows.

This is proved by two C-S or equivalents, so not the most strongest inequality. I would not be surprised the optimal value of such $p$ is strictly less than $2.$