Let $a+b+c=3,$ then prove$\sqrt[3]{\frac{a+b}{5ab+4}}+\sqrt[3]{\frac{c+b}{5cb+4}}+\sqrt[3]{\frac{a+c}{5ac+4}}\ge \sqrt[3]{6}$

Question

Problem

If $a,b,c\ge 0: a+b+c=3,$ then $$\sqrt[3]{\frac{a+b}{5ab+4}}+\sqrt[3]{\frac{c+b}{5cb+4}}+\sqrt[3]{\frac{a+c}{5ac+4}}\ge \sqrt[3]{6}.\tag{1}$$ I came up with when trying to prove $$\frac{a+b}{5ab+4}+\frac{c+b}{5cb+4}+\frac{a+c}{5ac+4}\ge \frac{2}{3},$$which is not hard by $uvw$ method. I'm looking for a nice proof $(1).$ It might be useful if the solver give motivate path.

Thank you for helping.

Answer 1

Some thoughts.

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Firstly, we'll prove the following inequality as a lemma.

Lemma. For all non-negative real numbers $x,y,z$ satisfying$$x^3y^3+y^3z^3+z^3x^3+\frac{537}{64}(xyz)^3\ge \frac{729}{64},$$then $$x+y+z\ge 3.$$ Equality holds at $(x,y,z)=(1,1,1);\left(0,\dfrac{3}{2},\dfrac{3}{2}\right)$ and permutations.

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Now, let $$x=\sqrt[3]{\frac{9}{2}}.\sqrt[3]{\frac{a+b}{5ab+4}};y=\sqrt[3]{\frac{9}{2}}.\sqrt[3]{\frac{c+b}{5cb+4}};z=\sqrt[3]{\frac{9}{2}}.\sqrt[3]{\frac{a+c}{5ac+4}}.$$The original problem is rewritten according to $x,y,z$ $$x+y+z\ge 3,$$which the lemma says that it's enough to prove $$\frac{81}{4}\sum_{cyc}\frac{(a+b)(b+c)}{(5ab+4)(5bc+4)}+\frac{391473}{512}\frac{(a+b)(b+c)(c+a)}{(5ab+4)(5bc+4)(5ca+4)}\ge \frac{729}{64}.$$ From now, we can use $uvw.$

About $uvw,$ see here.

Answer 2

Calling $f(a,b) = \sqrt[3]{\frac{a+b}{5ab+4}}$ and $F(a,b,c) = f(a,b)+f(b,c)+f(c,a)$ we have

$$ \cases{ f(a,b)\le F(a,b,c)\\ f(b,c)\le F(a,b,c)\\ f(c,a)\le F(a,b,c) } $$

so the minimum will be found amongst the problems

$$ \cases{ \min_{a,b}F(a,b,0),\ \text{s.t.}\ \ a+b=3,a \ge 0,b\ge 0\\ \min_{b,c}F(0,b,c),\ \text{s.t.}\ \ b+c=3,b \ge 0,c\ge 0\\ \min_{a,c}F(a,0,c),\ \text{s.t.}\ \ a+c=3,a \ge 0,c\ge 0\\ } $$

as the problem has nice symmetry we can follow with

$$ \min_{a,b}\sqrt[3]{\frac{a+b}{5ab+4}}+\sqrt[3]{\frac{b}{4}}+\sqrt[3]{\frac{a}{4}}\ \ \text{s.t.}\ \ \ a+b=3, a\ge 0, b\ge 0 $$

or

$$ \min_a\sqrt[3]{\frac{3}{5a(3-a)+4}}+\sqrt[3]{\frac{3-a}{4}}+\sqrt[3]{\frac{a}{4}}\ \ \text{s.t.}\ \ a \ge 0 $$

We have solutions at

$$ \cases{ a = 3, b = c = 0\\ a = 0, b = 3, c= 0\\ a = 0, b= 0, c=3\\ a = b = c = 1\\ } $$

with value $\sqrt[3]{6}$. The last solution cannot be detected by this procedure.

Answer 3

Some thought.

We use the isolated fudging.

It suffices to prove that $$\sqrt[3]{\frac{a + b}{6(5ab + 4)}} \ge \frac{55(a^2 + b^2) + 254ab + 385c(a + b)}{110(a^2 + b^2 + c^2) + 1024(ab + bc + ca)}.\tag{1}$$ (Note: Taking cyclic sum on (1), we get the desired inequality.)

(1) is true which is verified by Mathematica.

Answer 4

A full TATA box's solution.

Lemma.Let $x$, $y$ and $z$ be non-negative numbers such that: $$x^3y^3+x^3y^3+y^3z^3+\frac{537}{64}x^3y^3z^3\geq\frac{729}{64}.$$ Prove that: $$x+y+z\geq3.$$

Let $x+y+z<3,$ $x=kp$, $y=kq$ and $z=kr$, where $k>0$ and $p+q+r=3$.

Thus, $$kp+kq+kr<3=p+q+r,$$ which gives $k<1$ and we obtain: $$\frac{729}{64}\leq x^3y^3+x^3y^3+y^3z^3+\frac{537}{64}x^3y^3z^3<k^6(p^3q^3+p^3r^3+q^3r^3)+k^9\frac{537}{64}p^3q^3r^3<$$ $$<p^3q^3+p^3r^3+q^3r^3+\frac{537}{64}p^3q^3r^3,$$ which is a contradiction because we'll prove now that: $$p^3q^3+p^3r^3+q^3r^3+\frac{537}{64}p^3q^3r^3\leq\frac{729}{64}$$ for any non-negatives $p$, $q$ and $r$ such that $p+q+r=3$.

Indeed, let $p+q+r=3u$, $pw+pr+qr=3v^2$ and $pqr=w^3$.

Thus, $u=1$ and we need to prove that: $$27v^6-27uv^2w^3+3w^6+\frac{537}{64}w^9\leq\frac{729}{64}$$ or $f(w^3)\geq0,$ where $$f(w^3)=243u^9-179w^9-u^3(9v^6-9uv^2w^3+w^6).$$ Easy to see that $f$ is a concave function. But the concave function gets a minimal value for an extremal value of $w^3$,which by $uvw$ (https://artofproblemsolving.com/community/c6h278791) happens in the following cases.

1. $w^3=0$.

Let $r=0$.

Thus, by AM-GM $$p^3q^3+p^3r^3+q^3r^3+\frac{537}{64}p^3q^3r^3=p^3q^3\leq\left(\frac{p+q}{2}\right)^6=\frac{729}{64}.$$

2. Two variables are equal.

Let $q=p$ and $r=3-2p$, where $0\leq p\leq\frac{3}{2}.$

Thus, it remains to prove that: $$p^6+2p^3(3-2p)^3+\frac{537}{64}p^6(3-2p)^3\leq\frac{729}{64}$$ or $$(p-1)^2(3-2p)(81+216p+387p^2+198p^3+537p^4+716p^5-716p^6)\geq0,$$ which is true for $0\leq p\leq \frac{3}{2}.$

Id est, by using lemma, it's enough to prove that: $$128\sum_{cyc}\frac{(a+b)(a+c)}{(5ab+4)(5ac+4)}+4833\prod_{cyc}\frac{a+b}{5ab+4}\geq72.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $$128\sum_{cyc}(a+b)(a+c)(5bc+4)+4833\prod_{cyc}(a+b)\geq72\prod_{cyc}(5ab+4)$$ and we see again that we need to prove $g(w^3)\geq0,$ where $g$ is a concave function(the LHS is a linear of $w^3$ and a coefficient before $w^6$ in the RHS is equal to $72\cdot5^3>0$.

Thus, it's enough to check two following cases.

1. $c=0$.

We need to prove here $$128(3a\cdot4+3b\cdot4+ab(5ab+4))+4833\cdot3ab\geq72\cdot4\cdot4(5ab+4)$$ or $$128ab(5ab+4)+4833\cdot3ab\geq72\cdot4\cdot4\cdot5ab,$$ which is obvious.

2. $b=a$ and $c=3-2a,$ where $0\leq a\leq1.5$.

Here we obtain: $$(a-1)^2a(9251+5460a+6000a^2-6000a^3)\geq0,$$ which is true for $0\leq a\leq1.5$.

Answer 5

Let $\;f(x,y)=\sqrt[\Large 3]{\dfrac{x+y}{4+5x y}},\;$ then $$F(a, b, c)=f(a,b)+f(b,c)+f(c,a),$$ $$\dfrac{\partial}{\partial x}\,f(x,y) =\dfrac13 \dfrac{}{}\sqrt[\Large3]{\dfrac{5xy+4}{x+y}}^2\, \dfrac{4+5xy-5y(x+y)}{(4+5xy)^2} =\dfrac13\dfrac{4-5y^2}{\sqrt[\Large 3\,]{(4+5xy)^4(x+y)^2}\mathstrut} =g(x,y),$$ $$\dfrac{\partial}{\partial y}\,f(x,y) = g(y,x),$$

$\color{brown}{\textbf{The inner stationary points.}}$

The inner stationary points can be found from the system
$$\begin{cases} \dfrac{\partial}{\partial a}\,(F(a,b,c)+\lambda(3-a-b-c))=g(a,b)+g(a,c)-\lambda=0\\[4pt] \dfrac{\partial}{\partial b}\,(F(a,b,c)+\lambda(3-a-b-c))=g(b,a)+g(b,c)-\lambda=0\\[4pt] \dfrac{\partial}{\partial c}\,(F(a,b,c)+\lambda(3-a-b-c))=g(c,b)+g(c,a)-\lambda=0\\[4pt] \dfrac{\partial}{\partial \lambda}\,(F(a,b,c)+\lambda(3-a-b-c))=3-a-b-c=0. \end{cases}$$ or $$\begin{cases} \dfrac{4-5b^2}{\sqrt[\Large 3\,]{(4+5ab)^4(a+b)^2}\mathstrut} +\dfrac{4-5c^2}{\sqrt[\Large 3\,]{(4+5ca)^4(c+a)^2}\mathstrut}=\lambda\\[4pt] \dfrac{4-5a^2}{\sqrt[\Large 3\,]{(4+5ab)^4(a+b)^2}\mathstrut} +\dfrac{4-5c^2}{\sqrt[\Large 3\,]{(4+5bc)^4(b+c)^2}\mathstrut}=\lambda\\[4pt] \dfrac{4-5b^2}{\sqrt[\Large 3\,]{(4+5bc)^4(b+c)^2}\mathstrut} +\dfrac{4-5a^2}{\sqrt[\Large 3\,]{(4+5ca)^4(c+a)^2}\mathstrut}=\lambda\\[4pt] a+b+c=3 \end{cases}$$

After eliminating of $\lambda$ we get the overdefined system

$$\begin{cases} \dfrac{5(a^2-b^2)}{\sqrt[\Large 3\,]{(4+5ab)^4(a+b)^2}\mathstrut} +(4-5c^2)\left(\dfrac1{\sqrt[\Large 3\,]{(4+5ca)^4(c+a)^2}\mathstrut} -\dfrac1{\sqrt[\Large 3\,]{(4+5bc)^4(b+c)^2}\mathstrut}\right)=0\\[4pt] \dfrac{5(b^2-c^2)}{\sqrt[\Large 3\,]{(4+5bc)^4(b+c)^2}\mathstrut} +(4-5a^2)\left(\dfrac1{\sqrt[\Large 3\,]{(4+5ab)^4(a+b)^2}\mathstrut} -\dfrac1{\sqrt[\Large 3\,]{(4+5ca)^4(c+a)^2}\mathstrut}\right)=0\\[4pt] \dfrac{5(c^2-a^2)}{\sqrt[\Large 3\,]{(4+5ca)^4(c+a)^2}\mathstrut} +(4-5b^2)\left(\dfrac1{\sqrt[\Large 3\,]{(4+5bc)^4(b+c)^2}\mathstrut} -\dfrac1{\sqrt[\Large 3\,]{(4+5ab)^4(a+b)^2}\mathstrut}\right)=0\\[4pt] a+b+c=3, \end{cases}$$

If $\;a>b>c>0,\;$ then

• $\;(4+5ca)^4(c+a)^2 > (4+5bc)^4(b+c)^2\;$ and $\;4-5c^2>0.$
• $\;(4+5ab)^4(a+b)^2 > (4+5ca)^4(c+a)^2\;$ and $\;4-5a^2>0.$
• $\;(4+5ab)^4(a+b)^2 > (4+5ca)^4(c+a)^2\;$ and $\;4-5b^2>0.$

If $\;0<a<b<c,\;$ then

• $\;(4+5ca)^4(c+a)^2 < (4+5bc)^4(b+c)^2\;$ and $\;4-5c^2>0.$
• $\;(4+5ab)^4(a+b)^2 < (4+5ca)^4(c+a)^2\;$ and $\;4-5a^2>0.$
• $\;(4+5ab)^4(a+b)^2 < (4+5ca)^4(c+a)^2\;$ and $\;4-5b^2>0.$

Both of the variants mean that $\;a<1,b<1,c<1,\;$ i.e. $a+b+c<3,$ in contradiction with the last equation of the system.

If WLOG $\;a=b,\;$ then $$\begin{cases} \dfrac{5(a^2-c^2)}{\sqrt[\Large 3\,]{(4+5ac)^4(a+c)^2}\mathstrut} +(4-5a^2)\left(\dfrac1{\sqrt[\Large 3\,]{(4+5a^2)^4(2a)^2}\mathstrut} -\dfrac1{\sqrt[\Large 3\,]{(4+5ca)^4(c+a)^2}\mathstrut}\right)=0\\[4pt] \dfrac{5(c^2-a^2)}{\sqrt[\Large 3\,]{(4+5ca)^4(c+a)^2}\mathstrut} +(4-5a^2)\left(\dfrac1{\sqrt[\Large 3\,]{(4+5ac)^4(a+c)^2}\mathstrut} -\dfrac1{\sqrt[\Large 3\,]{(4+5a^2)^4(2a)^2}\mathstrut}\right)=0\\[4pt] 2a+c=3, \end{cases}$$

$$\begin{cases} \dfrac{10a^2-5c^2-4}{\sqrt[\Large 3\,]{(4+5ac)^4(a+c)^2}\mathstrut} +\dfrac{4-5a^2}{\sqrt[\Large 3\,]{(4+5a^2)^4(2a)^2}\mathstrut}=0\\[4pt] 2a+c=3, \end{cases}$$

$$(10a^2-5c^2-4)^3(4+5a^2)^4(2a)^2=(5a^2-4)^3(4+5ac)^4(a+c)^2,$$ $$(10a^2-5(3-2a)^2-4)^3(4+5a^2)^4(2a)^2=(5a^2-4)^3(4+5a(3-2a))^4(3-a)^2,$$ $$(-10a^2+60a-49)^3(4+5a^2)^4(2a)^2=(5a^2-4)^3(-10a^2+15a+4)^4(3-a)^2,$$ with the roots in $(0,3)$ $$\dbinom{a}{F(a,a,3-2a)}\in\left\{\approx\dbinom{0.05342847}{2.00156},\dbinom{1}{\sqrt[\Large3]6}\right\}.$$

Therefore, the least inner value of $F$ is $\sqrt[\Large 3]6.$

$\color{brown}{\textbf{The edges.}}$

Since $$\;F(3,0,0)=2\,\sqrt[\Large3]{\dfrac34}=\sqrt[\Large3]6$$ and $$\;F\left(\dfrac32,\dfrac32,0\right)=\sqrt[\Large3\,]{\dfrac3{4+\dfrac{45}4}} +2\sqrt[\Large3]{\dfrac38}=\sqrt[\Large3]{\dfrac{12}{61}}+\sqrt[\Large3]3 >\sqrt[\Large3]6,$$ then $$\color{green}{\mathbf{F(a,b,c)\ge \sqrt[\Large3]6}}$$ at the task constraints.