# If $x, y, z$ are real numbers with $xy+yz+zx\geq0,$ then $\left(x^5 + y^5 + z^5\right)^2 \geq3xyz\left(x^7 +y^7 + z^7\right).$

Question: If $$x, y, z$$ are real numbers with $$xy+yz+zx\geq0,$$ then

$$\left(x^5 + y^5 + z^5\right)^2 \geq3xyz\left(x^7 +y^7 + z^7\right).$$

Attempt:

Let's consider the expression $$\left(x^5 + y^5 + z^5\right)^2 \geq 3xyz\left(x^7 +y^7 + z^7\right)$$ and try to manipulate it.

We know that $$xy + yz + zx \geq 0,$$ which implies that the terms involving $$x$$, $$y$$, and $$z$$ are non-negative.

First we expand the left side of the inequality:

$$\left(x^5 + y^5 + z^5\right)^2 = x^{10} + y^{10} + z^{10} + 2(x^5y^5 + y^5z^5 + z^5x^5).$$

Now we aim to manipulate the right side:

$$3xyz\left(x^7 +y^7 + z^7\right) = 3xyz(x^7 + y^7 + z^7).$$

Here, we note that the terms $$x^7$$, $$y^7$$, and $$z^7$$ have greater powers compared to the terms $$x^5$$, $$y^5$$, and $$z^5$$ on the left side. Thus, we can't directly compare them.

We need to establish a connection between the two sides. However, the given condition $$xy + yz + zx \geq 0$$ doesn't directly help us in this context.

• If $x,y,z\ge 0$, it is a problem of AMM. Otherwise, WLOG, assume that $x, y \le 0$ and $z=1$. Then some BW (Buffalo Way) -like approach should work. But it is not nice. Commented May 13 at 10:27
• @RiverLi can you elaborate your proof? Commented May 13 at 12:12
• It is not nice. Perhaps in some days. Commented May 13 at 12:22
• @RiverLi any progress? Commented May 18 at 3:35
• I will post something soon. Commented May 18 at 5:20

As stated by River in a comment, for $$x,y,z>0$$ there is a proof for the inequality (see solution of AMM Problem 11543 in page 251 of this pdf), and the remaining case reduces to check the inequality for $$z=1, x,y<0$$, i.e, if $$xy+x+y>0$$, then

$$\left(1+x^5 + y^5\right)^2 \geq3xy\left(1+x^7 +y^7 \right).$$

Let us set $$a=-x> 0$$ and $$b=-y> 0$$. Then, we need to show for $$ab\ge a+b, a,b > 0$$:

$$\left(1-a^5 - a^5\right)^2 \geq3ab\left(1-a^7 -a^7 \right).$$

For $$1 \le a^7 +a^7$$, the above holds. For $$a^7 +a^7 < 1$$, we have $$0, but the initial condition $$ab\ge a+b \Leftrightarrow 1\ge \frac 1 a +\frac 1 b$$ cannot hold, and we are done.

• Nice. In my opinion, I did not see a nice proof for $x, y, z > 0$. Hope to see nice proof for $x, y, z > 0$. Commented May 19 at 9:54
• By the way, in your answer, alternatively, we can say $ab \ge a + b$ implies $a\ge 1, b \ge 1$. Commented May 19 at 9:56
• Thank you River! You are always supportive.
– Amir
Commented May 19 at 9:58

As @Amir's nice answer pointed out, the difficult case is $$x, y, z > 0$$ which is just AMM, 2010, no. 10, Problem 11543.

Here we give a proof of AMM Problem 11543 using the pqr method.

WLOG, assume that $$xy + yz + zx = 3$$. Let $$p = x + y + z, q = xy + yz + zx = 3, r = xyz$$. Using $$p^2 \ge 3q$$, we have $$p\ge 3$$.

The desired inequality is written as \begin{align*} f(r) &:= -21\,p{r}^{3}+ \left( 4\,{p}^{4}+39\,{p}^{2}+36 \right) {r}^{2} + \left( 7\,{p}^{7}-117\,{p}^{5}+522\,{p}^{3}-783\,p \right) r\\ &\qquad +{p}^{10} -30\,{p}^{8}+315\,{p}^{6}-1350\,{p}^{4}+2025\,{p}^{2} \ge 0.\tag{1} \end{align*}

We have $$0 \le (x - y)^2(y-z)^2(z-x)^2 = -4\,{p}^{3}r+{p}^{2}{q}^{2}+18\,pqr-4\,{q}^{3}-27\,{r}^{2}$$ which results in $$g(r) := -27\,{r}^{2}+ \left( -4\,{p}^{3}+54\,p \right) r+9\,{p}^{2}-108 \ge 0.\tag{2}$$

From (1) and (2), it suffices to prove that $$9 f(r) - 7pr g(r) \ge 0,$$ or \begin{align*} h(r) &:= \left( 64\,{p}^{4}-27\,{p}^{2}+324 \right) {r}^{2}+ \left( 63\,{p}^{7 }-1053\,{p}^{5}+4635\,{p}^{3}-6291\,p \right) r\\ &\qquad + 9p^2(p^4-15p^2+45)^2 \ge 0. \tag{3} \end{align*}

From (2), we have $$r \le r_2 := -\frac{2}{27}p^3 + p + \frac{2}{27}\sqrt{(p^2 - 9)^3}. \tag{4}$$

From (3) and (4), it suffices to prove that $$h(r) \ge 0$$ for all $$r\in [0, r_2]$$.

(i) If $$p \ge 59/18$$, we have $$63\,{p}^{7 }-1053\,{p}^{5}+4635\,{p}^{3}-6291\,p > 0$$, and thus $$h(r) \ge 0$$ for all $$r \ge 0$$.

(ii) If $$3 \le p \le 61/20$$, we have \begin{align*} h'(r) &= 2\left( 64\,{p}^{4}-27\,{p}^{2}+324 \right) r + \left( 63\,{p}^{7 }-1053\,{p}^{5}+4635\,{p}^{3}-6291\,p \right) \\ &\le 2\left( 64\,{p}^{4}-27\,{p}^{2}+324 \right) r_2 + \left( 63\,{p}^{7 }-1053\,{p}^{5}+4635\,{p}^{3}-6291\,p \right)\\ & < 0. \end{align*} Thus, $$h(r) \ge h(r_2) \ge 0$$.

(iii) If $$61/20 < p < 59/18$$, it suffices to prove that \begin{align*} &4 \times \left( 64\,{p}^{4}-27\,{p}^{2}+324 \right) \times 9p^2(p^4-15p^2+45)^2 \\ \ge{}& \left( 63\,{p}^{7 }-1053\,{p}^{5}+4635\,{p}^{3}-6291\,p \right)^2 \end{align*} which is true.

We are done.