let $a,b,c >0 $ and $abc=1$,prove that $\sqrt{1+8a^2}+ \sqrt{1+8b^2}+ \sqrt{1+8c^2}\leq 3(a+b+c )$ let $a,b,c >0 $ and $abc=1$,prove that $\sqrt{1+8a^2}+ \sqrt{1+8b^2}+ \sqrt{1+8c^2}\leq 3(a+b+c )$
can anyone help me with this question.
i've tried to assume that $a\geq b \geq c $ as my teacher said,however i couldn't solve it

WLOG: let $c=\min{(a,b,c)}$,then we have $c\le 1$, use Cauchy-Schwarz inequality we have
$$\sqrt{8a^2+1}+\sqrt{8b^2+1}\le\sqrt{(a+b)\left(\dfrac{8a^2+1}{a}+\dfrac{8b^2+1}{b}\right)}=(a+b)\sqrt{c+8}$$
we only prove
$$(a+b)(3-\sqrt{c+8})\ge\sqrt{8c^2+1}-3c$$
use AM-GM inequality we have
$$(a+b)(3-\sqrt{c+8})\ge 2\sqrt{ab}(3-\sqrt{c+8})=\dfrac{2(3-\sqrt{c+8}}{\sqrt{c}}$$
we only prove
$$\dfrac{6}{\sqrt{c}}+3\sqrt{c}\ge 2\sqrt{c+8}+\sqrt{8c^2+1}$$
It is easy prove it
