Prove $\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1$ for $ab + bc + ca = 3$

Question

Let $a,b,c\ge 0: ab+bc+ca=3.$ Prove that$$\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1.$$ This problem is from a book.

This cyclic inequality becomes an equality at $a=b=c=1.$ I try to used some estimate which based on this point $(1,1,1)$.

By AM-GM$$\sum_{cyc}\frac{1}{\sqrt{a+8b}}\ge 6\sum_{cyc}\frac{1}{a+8b+9}, $$but$$\frac{1}{a+8b+9}+\frac{1}{b+8c+9}+\frac{1}{c+8a+9}\ge \frac{1}{6}$$ is already wrong at $a=\dfrac{\sqrt{30}}{50},b=5\sqrt{30},c=0.$

The following Cauchy-Schwarz also doesn't help here since $$\sqrt{a+8b}+\sqrt{b+8c}+\sqrt{c+8a}\le 9,$$ is doesn't hold when $a=\dfrac{\sqrt{30}}{50},b=5\sqrt{30},c=0.$ Even $(a+8b)(b+8c)(c+8a)\le 27$ leads wrong inequality with same counter example.

Now, I tried Holder$$\left(\sum_{cyc}\frac{1}{\sqrt{a+8b}}\right)^2.\sum_{cyc}\left[(a+8b)(ma+nb+pc)^3\right]\ge \left[(m+n+p)(a+b+c)\right]^3,$$but it seems hard. I can not find any special $(m,n,p)$.

I think my Holder estimae is not good enough. Hope to see better ideas. Thank you.

Answer 1

Remarks: Here is an alternative proof. The proof of (1) is not nice by hand. (1) is true for all $b, c \in [0, 2]$. Hope to see a nice proof of (1).

WLOG, assume that $a = \max(a, b, c)$.

By AM-GM, we have $$\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}} \ge \frac{6}{a + 8b + 9} + \frac{2}{\frac{b + 8c}{2c + 1} + 2c + 1} + \frac{6}{c + 8a + 9}.$$

It suffices to prove that $$\frac{6}{a + 8b + 9} + \frac{2}{\frac{b + 8c}{2c + 1} + 2c + 1} + \frac{6}{c + 8a + 9} \ge 1.$$

Using $a = \frac{3 - bc}{b + c}$, after clearing the denominators, it suffices to prove that \begin{align*} f(b, c) &:= 224\,{b}^{3}{c}^{3}+164\,{b}^{2}{c}^{4}-28\,b{c}^{5}+56\,{b}^{4}c+393 \,{b}^{3}{c}^{2}+225\,{b}^{2}{c}^{3}-68\,b{c}^{4}\\ &\qquad -12\,{c}^{5}-24\,{b}^ {4}-80\,{b}^{3}c-704\,{b}^{2}{c}^{2}-368\,b{c}^{3}+96\,{c}^{4}-45\,{b} ^{3} -861\,{b}^{2}c\\ &\qquad -165\,b{c}^{2}+315\,{c}^{3}+300\,{b}^{2}+444\,bc-96 \,{c}^{2}+333\,b-171\,c+72\\ &\ge 0. \tag{1} \end{align*}

From $a = \max(a, b, c)$ and $ab + bc + ca = 3$, we have $b, c \le \sqrt 3 < 2$.

If $bc = 0$, it is easy to prove that (1) is true.

In the following, assume that $b, c > 0$.

Let $x = \frac{2}{b} - 1$ and $y = \frac{2}{c} - 1$. Then $x, y > 0$. We have \begin{align*} f(b, c) &= {\frac {6{x}^{2} \left( 12\,{y}^{5}+3\,{y}^{4}-172\,{y}^{3}+6\,{y}^{ 2}+736\,y+503 \right) \left( x-1 \right) ^{2}}{ \left( 1+x \right) ^{ 4} \left( 1+y \right) ^{5}}} \\[8pt] &\qquad + {\frac {2x \left( 549\,{y}^{5}+2607\,{y}^{4}+3126\,{y}^{3}+3842\,{y} ^{2}+9509\,y+4687 \right) \left( x-1 \right) ^{2}}{ \left( 1+x \right) ^{4} \left( 1+y \right) ^{5}}}\\[8pt] &\qquad + {\frac {2 [g(x, y)]^2 + 2h(x, y)(y - 1)^2}{ \left( 1+x \right) ^{4} \left( 1+y \right) ^{5} \left( 2877\,{y}^{5}+12474\,{y}^{4}+8930\,{y}^{3}-8984\,{y}^{2}+3617\,y+4190 \right) }}\\[6pt] &\ge 0 \end{align*} where \begin{align*} g(x, y) &:= 2877\,x{y}^{5}+12474\,x{y}^{4}+807\,{y}^{5}+8930\,x{y}^{3}+1330\,{y}^{ 4}-8984\,x{y}^{2}\\ &\qquad -5706\,{y}^{3}+3617\,xy-14048\,{y}^{2}+4190\,x-6301\, y+814, \end{align*} and \begin{align*} h(x, y) &:= 1066320\,{y}^{8}+8911776\,{y}^{7}+24116192\,{y}^{6}+24094752\,{y}^{5} + 40308096\,{y}^{4} \\ &\qquad +175656160\,{y}^{3}+304888864\,{y}^{2}+204549984\,y+ 44815664. \end{align*}

We are done.

Answer 2

The Contradiction method helps!

Indeed, let $\sum\limits_{cyc}\frac{1}{\sqrt{a+8b}}<1$,$a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $$\sum_{cyc}\frac{1}{\sqrt{x+8y}}=1.$$

Thus, $$\frac{1}{\sqrt{k}}\sum_{cyc}\frac{1}{\sqrt{x+8y}}<\sum_{cyc}\frac{1}{\sqrt{x+8y}},$$ which gives $k>1$ and $$3=ab+ac+bc=k^2(xy+xz+yz)>xy+xz+yz,$$ which is a contradiction because we'll prove now that $$xy+xz+yz\geq3$$ for any positives $x$, $y$ and $z$ such that $\sum\limits_{cyc}\frac{1}{\sqrt{x+8y}}=1.$

Now, let $\frac{1}{\sqrt{x+8y}}=\frac{p}{3},$ $\frac{1}{\sqrt{y+8z}}=\frac{q}{3}$ and $\frac{1}{\sqrt{z+8x}}=\frac{r}{3}.$

Thus, $$p+q+r=3,$$ $$x=\frac{\frac{1}{p^2}-\frac{8}{q^2}+\frac{64}{r^2}}{57}\geq0,$$ $$y=\frac{\frac{1}{q^2}-\frac{8}{r^2}+\frac{64}{p^2}}{57}\geq0$$ and $$z=\frac{\frac{1}{r^2}-\frac{8}{p^2}+\frac{64}{q^2}}{57}\geq0$$ and we need to prove that: $$\sum_{cyc}\left(\frac{1}{p^2}-\frac{8}{q^2}+\frac{64}{r^2}\right)\left(\frac{1}{q^2}-\frac{8}{r^2}+\frac{64}{p^2}\right)\geq3\cdot57^2$$ or $$\sum_{cyc}(65p^4q^2r^2-8p^4q^4)\geq171p^4q^4r^4,$$ which after substitution $p+q+r=3u$, $pq+pr+qr=3v^2$, $pqr=w^3$ gives $$65(9u^2-6v^2)u^4w^6-8(18u^2w^6+12v^2w^6-108uv^4w^3+81v^8)u^4\geq171w^{12}$$ or $f(w^3)\geq0,$ where $$f(w^3)=-19w^{12}+49u^6w^6-54u^2v^4w^6+96u^5v^4w^3-72u^4v^8.$$ But by Maclaurin: $$f'(w^3)=96u^5v^4+98u^6w^3-108u^2v^4w^3-76w^9>0,$$ which says that it's enough to prove $f(w^3)\geq0$ for the minimal value of $w^3$.

Now, $p$, $q$ and $r$ are positive roots of the equation: $$(t-p)(t-q)(t-r)=0$$ or $$t^3-3ut^2+3v^2t-w^3=0$$ or $g(t)=w^3$, where $$g(t)=t^3-3ut^2+3v^2t.$$

Let $u=constant$ and $v=constant$ and we want to move $w^3$.

During this moving should be that the equation $g(t)=w^3$ has three positive roots.

But $$g'(t)=3t^2-6ut+3v^2=3(t-t_1)(t-t_2),$$ where $t_1=u-\sqrt{u^2-v^2}$ and $t_2=u+\sqrt{u^2-v^2},$ which says $t_{max}=t_1$ and $t_{min}=t_2$.

Also, we have: $g(0)=0$ and we can draw a graph of $g$, which intersects with a line $y=w^3$ in three points or maybe, if this line is a tangent line to the graph of $g$, so they have two common points.

We see that $w^3$ gets a minimal value, when $y=w^3$ is a tangent line to a graph of $g$ in the point $(t_2,g(t_2))$. Also, we need to check, what happens for $w^3\rightarrow0^+$.

Id est, it's enough to prove $f(w^3)\geq0$ for equality case of two variables

(the case $w^3\rightarrow0^+$ is impossible because it should be $\sum\limits_{cyc}(65a^4b^2c^2-8a^4b^4)>0$).

Now, let $q=p$ and $r=3-2p$.

Thus, $$0<p<\frac{3}{2},$$ $$\frac{64}{(3-2p)^2}-\frac{7}{p^2}\geq0$$ and $$\frac{65}{p^2}-\frac{8}{(3-2p)^2}\geq0,$$ which gives $$\frac{4\sqrt7-7}{6}\leq p\leq\frac{65-\sqrt{130}}{42}$$ and we need to prove that $$130p^6(3-2p)^2+65p^4(3-2p)^4-8(p^8+2p^4(3-2p)^4)\geq171p^8(3-2p)^4$$ or $$(p-1)^2(441-294p+277p^2+152p^3-1368p^4+1216p^5-304p^6)\geq0,$$ which is true for $\frac{4\sqrt7-7}{6}\leq p\leq\frac{65-\sqrt{130}}{42}$.