# Inequality with a High Degree Constraint

This question-

Suppose that $$x, y, z$$ are positive real numbers and $$x^5 + y^5 + z^5 = 3$$. Prove that $${x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \ge 3$$

The inequality has a high degree constraint which can convert a $$5$$-degree polynomial to a $$0$$-degree term and makes it difficult.
On trying C-S to manage- $$\left({x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3}\right)\left(x^5 + y^5 + z^5\right) \ge 9 \Rightarrow \left(x^2y+y^2z+z^2x\right)^2\geq9 \Rightarrow x^2y+y^2z+z^2x\geq3$$Still gives a third degree inequality and not a useful fifth degree.

How can I do it and solve the problem?

Using the AM-GM inequality, we have $$\frac{30 x^4}{y^3} +7x^{10}+y^{10}+16x^5y^5 \geqslant 54\sqrt{\left(\frac{x^4}{y^3}\right)^{30} \cdot (x^{10})^7 \cdot y^{10} \cdot (x^5y^5)^{16}} = 54x^5.$$ Similar $$\frac{30 y^4}{z^3} +7y^{10}+z^{10}+16y^5z^5 \geqslant 54y^5,$$ and $$\frac{30 z^4}{x^3} +7z^{10}+x^{10}+16z^5x^5 \geqslant 54z^5.$$ Therefore $$30\left({x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \right) +8(x^5+y^5+z^5)^2 \geqslant 54(x^5+y^5+z^5).$$ So $${x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \geqslant .3$$

• How did you know that these values work the problem? Is there a technique for it? Aug 1, 2020 at 7:06
• @BookOfFlames Generally when you have to prove a sum is greater than something you tend to think of this inequality and then you know you need to form a 3 somehow so that's how you go ahead Aug 1, 2020 at 14:24

We'll prove that $$x^{6.5}+y^{6.5}+y^{6.5}\geq x^6+y^6+z^6,$$ for which it's enough to prove that: $$\sum_{cyc}\left(10x^{6.5}-10x^6-x^5+1\right)\geq0$$ or $$5\sum_{cyc}\left(2x^{1.5}-3x+1\right)x^{5}+\sum_{cyc}\left(5x^6-6x^5+1\right)\geq0,$$ which is true by AM-GM.

Now, by C-S and Vasc we obtain: $$\sum_{cyc}\frac{x^4}{y^3}=\sum_{cyc}\frac{x^{13}}{x^9y^3}\geq\frac{\left(\sum\limits_{cyc}x^{6.5}\right)^2}{\sum\limits_{cyc}x^9y^3}\geq\frac{\left(\sum\limits_{cyc}x^{6}\right)^2}{\sum\limits_{cyc}x^9y^3}\geq3.$$ The following inequality is also true.

Let $$x$$, $$y$$ and $$z$$ be positive numbers such that $$x^{34}+y^{34}+z^{34}=3.$$ Prove that: $$\frac{x^4}{y^3}+\frac{y^4}{z^3}+\frac{z^4}{x^3}\geq3$$

But it's not for contest.

• Why is the first inequality true? Aug 1, 2020 at 7:23
• @Pavel Kozlov I added something. I hope now it's clear. Aug 1, 2020 at 7:30
• That's very nice, thank you Aug 1, 2020 at 7:32
• @Pavel Kozlov You are welcome! Aug 1, 2020 at 7:33

A solution due to AoPS user @realquarterb:

By AM-GM, we have $$1 + 19 x^{100/19} \ge 20 x^5$$.

By AM-GM, we have $$10\frac{x^4}{y^3} + 3x^{10} + 6x^5y^5 \ge 19x^{100/19} \ge 20x^5 - 1$$.

Thus, we have $$10 \sum_{\mathrm{cyc}} \frac{x^4}{y^3} + 3(x^5+y^5+z^5)^2 \ge 20(x^5+y^5+z^5) - 3$$. (Q. E. D.)

• Why not even shorter: $10\frac{x^4}{y^3}+3x^{10}+6x^5y^5+1\ge 20x^5$ by AM-GM? Aug 1, 2020 at 14:23
• @Chrystomath Nice observation. Aug 1, 2020 at 14:43