Proving $\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}$ I've been going through some of my notes when I found the following inequality for $a,b,c>0$ and $abc=1$:
$$
\begin{equation*}
\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}
\end{equation*}
$$
This was what I attempted, but that yielded no result whatsoever:
$$
\begin{align}
\sqrt{2}(a+b+c) &\geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}\\
&\geq \sqrt{abc+a²} + \sqrt{abc+b²} + \sqrt{abc+c²}\\
&\geq \sqrt{a(bc+a)}+\sqrt{b(ac+b)}+\sqrt{c(ab+c)}\\
&\geq \sqrt{a\left(\frac{1}{a}+a\right)}+\sqrt{b\left(\frac{1}{b}+b\right)}+\sqrt{c\left(\frac{1}{c}+c\right)}
\end{align}
$$
With this we then have
$$
\begin{align}
\sqrt{1+a²} &= \sqrt{a\left(\frac{1}{a}+a\right)}\\
&=\sqrt{a\left(\frac{1}{a}+\frac{a²}{a}\right)}\\
&=\sqrt{a\left(\frac{1+a²}{a}\right)}\\
&=\sqrt{1+a²}
\end{align}
$$
This yields nothing but frustration and gets me back to step 0

here is another way which is not use derivative:
first we need to know: $\sqrt{\dfrac{x+y+z}{3}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}$ (1)
to prove it, we have  $\sqrt{\dfrac{x+y}{2}} \ge \dfrac{\sqrt{x}+\sqrt{y}}{2} \implies \sqrt{\dfrac{x+y+z+t}{4}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{t}}{4}$
 let $t= \dfrac{x+y+z}{3}$ we have (1)
squre both sides:
$2(a+b+c)^2 \ge 3+(a^2+b^2+c^2)+2\sum {\sqrt{1+a^2+b^2+a^2b^2}} \iff \sum a^2+4\sum ab -3 \ge 2\sum {\sqrt{1+a^2+b^2+a^2b^2}} \iff \sum a^2+2\sum ab -3 \ge 2\sqrt{3(3+2\sum a^2+\sum a^2b^2)} \iff (\sum a^2-3)^2+8(\sum a^2-3)(\sum ab)+16(\sum ab)^2 \ge 36+24\sum a^2+12 \sum a^2b^2 $
note $(\sum ab)^2-\sum a^2b^2=2abc(a+b+c)=2(a+b+c) \iff (\sum a^2-3)^2+8(\sum a^2)(\sum ab-3) +4((\sum ab)^2-6\sum ab+9)+24(a+b+c)-72 \ge 0$
$abc=1$, so it is trivial $a+b+c \ge 3, ab+bc+ac \ge 3$ so the last one is true.
