<s>[INST] 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
 [/INST] Let
$$ f(a,b,c)=3\sum_{cyc}a-\sum_{cyc}\sqrt{1+8a^2}.$$
Assuming that $a\geq b\geq c$, the Cauchy-Schwarz inequality gives:
$$ f(a,b,c)\geq f(\sqrt{ab},\sqrt{ab},c)\tag{1}$$
hence it is sufficient to prove the inequality in the case $(a,b,c)=\left(x,x,\frac{1}{x^2}\right):$
$$ 3\left(2x+\frac{1}{x^2}\right)-2\sqrt{1+8x^2}-\sqrt{1+\frac{8}{x^4}}\geq 0,$$
$$ 3(2x^3+1)\geq 2x^2\sqrt{1+8x^2}+\sqrt{x^4+8}$$
$$ 1+36x^3-5x^4+4x^6 \geq 4x^2\sqrt{(1+8x^2)(8+x^4)}.\tag{2}$$
To prove $(2)$ it is sufficient to prove that for any $x\in\mathbb{R}^+$ we have:
$$ 1+72x^3 - 138x^4 + 2328 x^6 - 360 x^7 \geq 0 \tag{3}$$ 
or the still weaker:
$$ 12-23x+328x^3 \geq 0,$$
$$ f(x)=1-2x+27x^3 \geq 0,\tag{4}$$
that is trivial since that cubic polynomial has a negative discriminant, hence only one real root, and since $f(0)>0$ while $f(-1)<0$ that root is between $-1$ and $0$, so the cubic polynomial is positive over $\mathbb{R}^+$.
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