Proved : $3(a+b+c)\geq \sqrt{8a^{2}+1}+\sqrt{8b^2+1}+\sqrt{8c^{2}+1}$

Question

$1$. $x;y;z\in \mathbb{R}$ such that $xyz=1$. Find the minimum or maximum value of : $\sum \dfrac{1}{x+1}$

$2$. $x;y;z\in \mathbb{R}^+$ such that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Proved : $3(a+b+c)\geq \sqrt{8a^{2}+1}+\sqrt{8b^2+1}+\sqrt{8c^{2}+1}$

Thanks :)

Answer 1

For 1)
Let $x=\epsilon-1, y = \frac1x, z = 1$, then we have $xyz=1$. Then $$ \sum \frac1{x+1} = - \frac1{\epsilon}- \frac{1-\epsilon}{2-\epsilon}+\frac12$$ Now by choosing $\epsilon \to 0^+$, we can make the sum arbitrarily small. OTOH, by letting $\epsilon \to 0^-$ we can make the sum arbitrarily large. Hence if $x, y, z \in \mathbb{R}$, i.e. negative numbers are allowed, there is no maximum or minimum.

For 2)
Let $f(t) = 3t-\sqrt{8t^2+1}-\dfrac16\left(t-\frac1t\right)$. Then the inequality is $f(a) + f(b)+f(c )\ge 0$. Hence it is sufficient to show $f(t)\ge 0$.

With $t = \dfrac{\tan \theta}{\sqrt8}$, it is equivalent to
$$8 \sqrt{2} \cot \theta-24 \sec \theta + 17 \sqrt{2} \tan \theta \ge 0$$ $$\iff 9\sqrt2 \left(\sin \theta - \frac{2\sqrt2}{3}\right)^2 \ge 0$$

Answer 2

for (1), if $x,y,z$ is positive,and such $xyz=1$ then we have $$1<\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}<2$$

since $$\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}=\dfrac{3+2(x+y+z)+xy+yz+xz}{2+xy+yz+xz+x+y+z}>1$$

and other hand $$\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}<\dfrac{4+2(x+y+z)+2(xy+yz+xz)}{2+xy+yz+xz+x+y+z}=2$$