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Let $x,y,z$ be positive real number such that $xy+yz+zx=3$. Find the minimum value of the expression $$P=\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}$$

Jimmy R.
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Phi Linh
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  • What have you tried? Because I will say that this is a fairly straightforward Lagrange multipliers problem, besides the fact that computing $\nabla P$ is rather tedious. – Ian Nov 17 '14 at 14:20
  • That my tried :3 – Phi Linh Nov 17 '14 at 14:21
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    I think $\frac{x^2}{\sqrt{x^3+8}}\ge \frac{2x^2}{x^2-2x+4+x+2}$ – Phi Linh Nov 17 '14 at 14:23
  • Not sure what that has to do with anything. Compute $\nabla P$ and $\nabla g$ where $g(x,y,z)=xy+yz+zx$ and solve $\nabla P=\nabla g$ coupled to the constraint. Equivalently, find a critical point of $L(x,y,z,\lambda)=P(x,y,z)-\lambda g(x,y,z)$. – Ian Nov 17 '14 at 14:31
  • ???? I don't understand – Phi Linh Nov 17 '14 at 14:35
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    @Ian Is it really that easy to compute? If you do the derivatives, you get an extremely ugly system of 4 equations and that helps exactly how? I don't see how is it easier to solve than this. – user2345215 Nov 17 '14 at 15:14

1 Answers1

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It is easy to guess from the symmetry that $x=y=z=1$ should give the minimum of $1$, and we will try proving this. By Cauchy-Schwarz Inequality:

$$\left(\sum_{cyc} \frac{x^2}{\sqrt{x^3+8}} \right)\left(\sum_{cyc} \sqrt{x^3+8} \right) \ge (x+y+z)^2 $$

We will use $\sum $ to denote cyclic sums. From the above, if we show that $(x+y+z)^2 \ge \sum \sqrt{x^3+8} $, then we have $P \ge 1$.

As you noted in comments, we can use $2\sqrt{x^3+8} \le (x+2)+(x^2-2x+4) = x^2-x+6$, so it is enough to show that: $$2\sum x^2+4\sum xy \ge \sum x^2-\sum x+18 \iff \sum x^2+\sum x \ge 6$$

But $\sum x^2 \ge \sum xy = 3$ and $\left( \sum x\right)^2 \ge 3\sum xy = 9 \implies \sum x \ge 3$, so this is true.

Macavity
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