How to solve this inequality, with the hypothesis more complicated than the conclusion?

Question

Given $x,y,z \in \mathbb{R}$ and $x,y,z>2,$ I want to show that if,

$$\frac{1}{x^2-4}+\frac{1}{y^2-4}+\frac{1}{z^2-4} = \frac{1}{7}$$

then,

$$\frac{1}{x+2} + \frac{1}{y+2} + \frac{1}{z+2} \leq \frac{3}{7}.$$

I follow the solution here: http://artofproblemsolving.com/community/c6h514107_inequality_by_poru_loh

but I don't know how to alter it to fit this problem?!

Answer 1

Just another way is to note that for $t> 2$, $$f(t) = \left(\frac17-\frac1{t+2}\right)-\frac9{10}\left(\frac1{21}-\frac1{t^2-4} \right)=\frac{(t-5)^2}{10(t^2-4)} \ge 0$$ and the inequality is equivalent to $f(x)+f(y)+f(z) \ge 0$. So its true and equality is iff $x=y=z=5$.

Answer 2

Here is an adaptation of hyperbolictangent's answer in your given link.

Let $S:= \frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}$. Note that \begin{align*} 3-4S &= \frac{x^2-4}{x^2-4} + \frac{y^2-4}{y^2-4} + \frac{z^2-4}{z^2-4} - 4 \left(\frac{x-2}{x^2-4}+\frac{y-2}{y^2-4}+\frac{z-2}{z^2-4}\right)\\ &= \frac{(x-2)^2}{x^2-4}+\frac{(y-2)^2}{y^2-4}+\frac{(z-2)^2}{z^2-4}\\ &= \frac{x-2}{x+2} + \frac{y-2}{y+2} + \frac{z-2}{z+2}. \end{align*}

Thus, \begin{align*} \frac{1}{7} \cdot (3-4S) &= \left(\frac{1}{x^2-4}+\frac{1}{y^2-4}+\frac{1}{z^2-4}\right) \left(\frac{x-2}{x+2} + \frac{y-2}{y+2} + \frac{z-2}{z+2}\right)\\ &\ge S^2 & \text{Cauchy-Schwarz} \end{align*}

So we have $$0 \ge S^2+\frac{4}{7}S-\frac{3}{7} = \left(S+\frac{2}{7}\right)^2 -\frac{25}{49}$$ $$\frac{5}{7} \ge S+\frac{2}{7}$$ $$\frac{3}{7} \ge S$$

Answer 3

Here is an adaptation of pi37 answer in your given link.

Note $$\sum_{cyc}\dfrac{x^2+25}{x^2-4}=\sum_{cyc}\dfrac{x^2-4+29}{x^2-4}=3+\dfrac{29}{7}=\dfrac{50}{7}$$ and use AM-GM $x^2+25\ge 10x$.so $$\sum_{cyc}\dfrac{x^2+25}{x^2-4}\ge\dfrac{10x}{x^2-4}\Longrightarrow\sum_{cyc}\dfrac{x}{x^2-4}\le \dfrac{5}{7}$$ and note $$\sum_{cyc}\dfrac{x}{x^2-4}-\sum_{cyc}\dfrac{2}{x^2-4}=\sum_{cyc}\dfrac{x-2}{x^2-4}=\sum_{cyc}\dfrac{1}{x+2}$$ so $$\sum_{cyc}\dfrac{1}{x+2}\le\dfrac{5}{7}-\dfrac{2}{7}=\dfrac{3}{7}$$