# 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

• can you tell us about your background in inequalities(you know about which inequalities)?are you familiar with AM-GM inequality?what about Cauchy schwarz inequality? – user2838619 Aug 3 '14 at 16:06
• $abc=1$ implies the existence of $x,y,z$ such that $a=x/y,b=y/z,c=z/x$.You could try substitution. – rah4927 Aug 3 '14 at 16:27
• You can find a proof here artofproblemsolving.com/Forum/viewtopic.php?f=151&t=597756 – rezvane Aug 3 '14 at 17:35

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 $\mathbb{R}^+$.

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

Let $a=e^x$, $b=e^y$ and $c=e^z$.

Hence, $x+y+z=0$ and we need to prove that $\sum\limits_{cyc}f(x)\geq0$, where

$f(x)=3e^x-\sqrt{1+8e^{2x}}$.

But $f''(x)>0$ for all $x>0$.

Thus, by Vasc's RCF Theorem it's enough to prove our inequality for $y=x$ and $z=-2x$

or the starting inequality for $b=a$ and $c=\frac{1}{a^2}$ and the rest is smooth.