Prove $\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\geq \sqrt{6x+6y+6z}$

Question

Prove, that for all positive $x,y,z$ following inequality holds $$\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\geq \sqrt{6x+6y+6z}$$

I have attempted several approaches to prove this inequality, including using Jensen's inequality and various elementary inequalities like $x^2+1\geq 2x$. However, none of these approaches led to the desired results. When using AM-GM inequality, I obtain a sixth root, which leads to nowhere.

Can someone provide me with hints or suggestions on how to approach this type of inequality?

Answer 1

By AM-GM, we have $$\sqrt{(x^2 + 1)(y^2 + 1)} = \sqrt{x^2y^2 + x^2 + y^2 + 1} \ge \sqrt{x^2y^2 + 2xy + 1} = xy + 1.$$ Similarly, we have $\sqrt{(y^2 + 1)(z^2 + 1)} \ge yz + 1$, and $\sqrt{(z^2 + 1)(x^2 + 1)} \ge zx + 1$.
(Note: Alternatively, we may use the Cauchy-Bunyakovsky-Schwarz inequality.)

Thus, we have \begin{align*} \mathrm{LHS}^2 &\ge x^2 + y^2 + z^2 + 3 + 2(xy + 1) + 2(yz + 1) + 2(zx + 1)\\ &= (x + y + z)^2 + 9. \end{align*}

Thus, we have $$\mathrm{LHS}^2 - \mathrm{RHS}^2 \ge (x + y + z)^2 + 9 - (6x + 6y + 6z) = (x + y + z - 3)^2 \ge 0.$$

We are done.

Answer 2

Hint: Define the funtion $f(t)= \sqrt{t^2 +1}$ which is convex ($f''(t) >0$), apply the Jensen's inequality $$f(x)+f(y)+f(z) \ge 3 f\left( \frac{x+y+z}{3} \right)$$ and finally use the fact that $$\sqrt{\left(\frac{x+y+z}{3} \right)^2 +1} \ge \sqrt{2\cdot\frac{x+y+z}{3}}$$