I'm trying to prove that $$ F(u)= \int_{0}^{1}\sqrt{(1+u'^2_x)}dx$$ is a convex function of $u=u(x)$ ; however after squaring both side twice of $$\sqrt{(1+(d(tu_1)/dx)^2)}+\sqrt{(1+(d((1-t)u_2)/dx)^2)} >\sqrt{(1+(d(tu_1+(1-t)u_2)/dx)^2)} $$I got so many terms which does lead to any useful result. Can anyone give me some hint?
(u is a function of x but may not be differentiable)
As you noted in the comments, just observe that
\begin{eqnarray*} \frac{d^2}{dx^2} (1 + x^2)^{1/2} & =& \frac{d}{dx} \frac{1}{2} \cdot 2x \cdot (1 + x^2)^{-1/2} \\ &= & \frac{d}{dx} x \cdot (1 + x^2)^{-1/2} \\ & =& (1+x^2)^{-1/2} + x \cdot (-1/2) \cdot 2x \cdot (1+x^2)^{-3/2}\\ & = & (1+x^2)^{-1/2} \cdot [1 - \frac{x^2}{1+x^2}] \\ &= &(1+x^2)^{-1/2} \cdot \frac{1}{1+x^2} > 0. \end{eqnarray*}
Hence, $G : \Bbb{R} \to \Bbb{R}, x \mapsto (1+x^2)^{1/2}$ is convex.
Using the monotonicity of the integral, this translates to $F$.
Indeed, for $\lambda \in [0,1]$, we have
\begin{eqnarray*} F(\lambda u + (1-\lambda)v) &=& \int_0^1 G(\frac{d}{dx} (\lambda u(x) + (1-\lambda) v(x))) \, dx \\ &=& \int_0^1 G(\lambda u'(x) + (1-\lambda) v'(x)) \,dx \\ & \leq & \int_0^1 \lambda G(u'(x)) + (1-\lambda) G(v'(x))\, dx &=& \lambda F(u) + (1-\lambda) F(v). \end{eqnarray*}
Still, $u$ and $v$ have to be differentiable for $F(u)$ and $F(v)$ to make sense.