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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?

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3 Answers 3

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

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    $\begingroup$ How did you know that these values work the problem? Is there a technique for it? $\endgroup$ Aug 1, 2020 at 7:06
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    $\begingroup$ @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 $\endgroup$ Aug 1, 2020 at 14:24
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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.

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

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

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