$a,b,c\ge 0: a+b+c+abc=4$. Prove that: $\sum_{cyc}\sqrt{\frac{a+b}{c+ab}}\ge 3.$

Question

Problem. Let $a,b,c\ge 0: a+b+c+abc=4$. Prove that: $$\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{b+c}{a+bc}}+\sqrt{\frac{c+a}{b+ca}}\ge 3.$$

I think the problem is nice and very hard. Equality holds iff $a=b=c=1; (0,2,2).$

The idea of using AM-GM directly doesn't work, because $$(a+b)(b+c)(c+a)\ge (a+bc)(b+ca)(c+ab),$$is wrong when $a=b=\dfrac{3}{2}.$

Also: $$\sum_{cyc}\sqrt{\frac{a+b}{c+ab}}\ge \sum_{cyc}\frac{2(a+b)}{a+b+c+ab},$$leads to wrong inequality.

Is Holder inequality is a good way in this case ?

I tried $$\left(\sum_{cyc}\sqrt{\frac{a+b}{c+ab}}\right)^2.\sum_{cyc}(a+b)^2(c+ab)\ge 8(a+b+c)^3,$$which is not good enough.

I think Mixing variables is a good approach but I am not good at using this method.

A fact: $$\sqrt[3]{\frac{a+b}{c+ab}}+\sqrt[3]{\frac{b+c}{a+bc}}+\sqrt[3]{\frac{c+a}{b+ca}}\ge 3,$$is already wrong when $a=b=1.874$.

Hope to see more ideas. Thank you.

Answer 1

Remarks: Since no verified proof is available, I give a proof. It is a computer assisted proof. It is very complicated by hand. Hope to see nice proofs.

Sketch of a proof.

Let $$x := \frac{a + b}{c + ab}, \quad y := \frac{b + c}{a + bc}, \quad z := \frac{c + a}{b + ca},$$ $$u := \frac{2(a + b)}{ab + a + b + c}, \quad v := \frac{2(b + c)}{bc + a + b + c}, \quad w := \frac{2(c + a)}{ca + a + b + c}.$$

By AM-GM, we have $u = \frac{2x}{1 + x} \le \sqrt x$, $v = \frac{2y}{1+y} \le \sqrt{y}$, and $w = \frac{2z}{1+z}\le \sqrt{z}$.

We have $$\sqrt{x} - u = \frac{x - u^2}{\sqrt{x} + u} \ge \frac{x - u^2}{\frac{x + 1}{2} + u}. \tag{1}$$

Using (1), it suffices to prove that $$u + \frac{x - u^2}{\frac{x + 1}{2} + u} + v + \frac{y - v^2}{\frac{y + 1}{2} + v} + w + \frac{z - w^2}{\frac{z + 1}{2} + w} \ge 3. \tag{2}$$

We use pqr method.

Let $p = a + b + c, q = ab + bc + ca, r = abc$.

The condition $a + b + c + abc = 4$ is written as $$p + r = 4. \tag{3}$$

(2) is written as \begin{align*} &5\,{p}^{6}+10\,{p}^{5}q-62\,{p}^{5}r+{p}^{4}{q}^{2}-14\,{p}^{4}qr-3\,{ p}^{4}{r}^{2}+24\,{p}^{4}q\\ &\quad +144\,{p}^{4}r-16\,{p}^{3}{q}^{2}+48\,{p}^{3 }qr-126\,{p}^{3}{r}^{2}-8\,{p}^{2}{q}^{3}-8\,{p}^{2}{q}^{2}r\\ &\quad -30\,{p}^{ 2}q{r}^{2}-6\,{p}^{2}{r}^{3}-96\,{p}^{3}r+16\,{p}^{2}{q}^{2}+288\,{p}^ {2}qr+252\,{p}^{2}{r}^{2}\\ &\quad -224\,p{q}^{3}+384\,p{q}^{2}r+72\,pq{r}^{2}- 64\,p{r}^{3}-48\,{q}^{4}+64\,{q}^{3}r+8\,{q}^{2}{r}^{2}\\ &\quad -16\,q{r}^{3}-3 \,{r}^{4}-256\,pqr-352\,p{r}^{2}+128\,{q}^{2}r-320\,q{r}^{2}+96\,{r}^{ 3}+192\,{r}^{2}\\ &\ge 0. \tag{4} \end{align*}

From (3), using $r = 4 - p$, (4) is written as \begin{align*} &-48\,{q}^{4}+ \left( -8\,{p}^{2}-288\,p+256 \right) {q}^{3}+ \left( {p }^{4}-8\,{p}^{3}-392\,{p}^{2}+1344\,p+640 \right) {q}^{2}\\ &\quad + \left( 24\, {p}^{5}-110\,{p}^{4}+232\,{p}^{3}-160\,{p}^{2}+3456\,p-6144 \right) q \\ &\quad + 64\,{p}^{6}-488\,{p}^{5}+1873\,{p}^{4}-5296\,{p}^{3}+10592\,{p}^{2}- 15104\,p+8448\\ &\ge 0. \tag{5} \end{align*}

From (3), we have $3 \le p \le 4$.

Using $q^2 \ge 3pr$ and (3), we have $q \ge \sqrt{3p(4 - p)}$.

Using degree four Schur inequality $r \ge \frac{5p^2q - p^4 - 4q^2}{6p}$ and (3), we have $$4 - p \ge \frac{5p^2q - p^4 - 4q^2}{6p}$$ which results in (using $p^2 \ge 3q$) $$q \le \frac58 p^2 - \frac18\sqrt{9p^4 + 96p^2 - 384p}.$$

Thus, we have $$3 \le p \le 4, \quad \sqrt{3p(4 - p)} \le q \le \frac58 p^2 - \frac18\sqrt{9p^4 + 96p^2 - 384p}. \tag{6}$$

If $q = \sqrt{3p(4 - p)}$, we can prove (5).

If $\sqrt{3p(4 - p)} < q \le \frac58 p^2 - \frac18\sqrt{9p^4 + 96p^2 - 384p}$, we use the Buffalo Way (BW).

Let $$q_1 := \sqrt{3p(4 - p)}, \quad q_2 := \frac58 p^2 - \frac18\sqrt{9p^4 + 96p^2 - 384p}.$$

Let $$s := \frac{q_2 - q}{q - q_1} \ge 0.$$ We have $$q = q_1 \cdot \frac{s}{1 + s} + q_2 \cdot \frac{1}{1 + s}. \tag{7}$$

We plug (7) into (5), after clearing the denominators, it suffices to prove that $$m_4s^4 + m_3s^3 + m_2s^2 + m_1s + m_0 \ge 0 \tag{8}$$ where $m_0, m_1, m_2, m_3, m_4$ are functions in $p$.

We can prove that $m_0, m_1, m_2, m_3, m_4 \ge 0$ for all $3\le p\le 4$. Thus, (8) is true.

We are done.

Answer 2

Some thoughts.

By Holder, we have $$\left(\sum_{\mathrm{cyc}}\sqrt{\frac{a+b}{c+ab}}\right)^2 \sum_{\mathrm{cyc}} (a + b)^2(c + ab)(c + 2)^3 \ge{} \left(\sum_{\mathrm{cyc}} (a + b)(c + 2)\right)^3.$$

It suffices to prove that $$\left(\sum_{\mathrm{cyc}} (a + b)(c + 2)\right)^3 \ge 9 \sum_{\mathrm{cyc}} (a + b)^2(c + ab)(c + 2)^3. \tag{1} $$ This inequality is true which is verified by Mathematica. It can be proved by the pqr method.

Let $p = a + b + c, q = ab + bc + ca, r = abc$.

The condition $a + b + c + abc = 4$ becomes $p + r = 4$.

Using $r = 4- p$, (1) is written as \begin{align*} &208\,{q}^{3}+ \left( -72\,{p}^{2}+96\,p-2304 \right) {q}^{2}+ \left( 480\,{p}^{2}-5616\,p+15552 \right) q\\ &\quad +2384\,{p}^{3}-14184\,{p}^{2}+ 34560\,p-31104 \ge 0. \end{align*}

Omitted.

Answer 3

Let $g(a,b,c)=\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{b+c}{a+bc}}+\sqrt{\frac{c+a}{b+ca}}$ and $f(a,b)=g(a,b,\frac{4-a-b}{1+ab})$ defined on the closed triangle except and with vertices $(0,0)$,$(4,0),(0,4)$. On boundry $f(a,b)$ reduces to the function $h(x)=\sqrt{\frac{4}{x(4-x)}}+\sqrt{\frac{x}{4-x}}+\sqrt{\frac{4-x}{x}}$ where $0< x< 4$ and by symmetry principle its minumum is $3$ at $x=2$. On the interior of the triangle, by symmetry principle, minumum occurs when $a=b=x$ and the problem reduces to finding the minumum of $$p(x)=\sqrt{\frac{2x^3+2x}{x^4+x^2-2x+4}}+2\sqrt{\frac{x^3-x+4}{x^3-2x^2+5x}}$$ where $0<x<2$. And the minimum is $3$ at $x=1$. See the link.

Answer 4

Just few remarks :

We can plays without loss of generality with the constraint we have :

$$a+b+c+abc=4$$

Now plugging $$a=y(u-v)\geq 0,b=y(v-w)\geq 0,c=z(u-w)\geq 0$$

Then we can factor this constraint as we have :

$$a+b+c+abc=y(u-v)+y(v-w)+z(u-w)+y(u-v)\cdot y(v-w)\cdot z(u-w)=\left(u-w\right)\left(uvy^{2}z-uwy^{2}z-v^{2}y^{2}z+vwy^{2}z+y+z\right)=4$$

On the other hand the inequality becomes :

$\sqrt{\frac{y\left(u-w\right)}{z\left(u-w\right)+y^{2}\left(u-v\right)\left(v-w\right)}}+\sqrt{\frac{y\left(v-w\right)+z\left(u-w\right)}{y\left(u-v\right)+y\left(v-w\right)z\left(u-w\right)}}+\sqrt{\frac{z\left(u-w\right)+y\left(u-v\right)}{y\left(v-w\right)+z\left(u-w\right)y\left(u-v\right)}}\ge 3$

Trivial Case : $u\geq 1=v=w$ :

A bit of algebra plugging $u\to ux,v\to vx,w\to wx,y\to yx,z\to zx$and we have :

$$a+b+c+abc=4=(u-1)x^2(y+z)\tag{I}$$

Again a bit of algebra and we need to show :

$$\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{y}}+\sqrt{\frac{z+y}{x^{2}zy\left(u-1\right)}}\ge 3$$

Or using $I$

$$\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{y}}+\sqrt{\frac{4}{x^{3}zy\left(u-1\right)^{2}}}\geq 3$$

Now $y\to y^2,x\to x^2,z\to z^2,u\to u+1$ we need to show :

$$\frac{y}{z}+\frac{z}{y}+\frac{2}{x^{3}zyu}\geq 3$$

Or :

$$\frac{\left(ux^{3}y^{2}-3ux^{3}yz+ux^{3}z^{2}+2\right)}{ux^{3}yz}\geq 0$$

Or :

$$\frac{\left(4x-3ux^{3}yz+2\right)}{ux^{3}yz}\geq 0$$

But $y^2+z^2\geq 2yz$ so we need to show :

$$\left(4x-\frac{3}{2}x+2\right)\geq 0$$

Wich is true !

Case : $u-v=v-w=1/2(u-w),\exists \varepsilon x,y,z\in[1+\varepsilon ,\infty),y\geq z$ (with the same substitution as in the trivial case)

Then we have :

$$a+b+c+abc=4=x^2(x^3y^2z+y+z)(u-w)$$

Then starting from :

$\sqrt{\frac{y\left(u-w\right)}{z\left(u-w\right)+x^2y^{2}\left(u-v\right)\left(v-w\right)}}+\sqrt{\frac{y\left(v-w\right)+z\left(u-w\right)}{y\left(u-v\right)+x^2y\left(v-w\right)z\left(u-w\right)}}+\sqrt{\frac{z\left(u-w\right)+y\left(u-v\right)}{y\left(v-w\right)+x^2z\left(u-w\right)y\left(u-v\right)}}\ge 3$

Or :

$$\sqrt{\frac{y\left(u-w\right)}{z\left(u-w\right)+x^2y^{2}\left(u-v\right)\left(v-w\right)}}+2\sqrt{\frac{y\left(v-w\right)+z\left(u-w\right)}{y\left(u-v\right)+x^2y\left(v-w\right)z\left(u-w\right)}}\ge 3$$

Or :

$$\sqrt{\frac{2y}{2z+x^2y^{2}\left(u-v\right)}}+2\sqrt{\frac{y+2z}{y\left(u-v\right)+2x^2y\left(v-w\right)z}}\ge 3$$

Or :

$$\sqrt{\frac{2y}{2z+2\frac{x^2y^2}{x^2(x^3y^2z+y+z)}}}+2\sqrt{\frac{y+2z}{y+4\frac{x^2yz}{x^2(x^3y^2z+y+z)}}}\ge 3$$

Or :

$$\sqrt{\frac{2y}{2z+2\frac{y^2}{(x^3y^2z+y+z)}}}+2\sqrt{\frac{y+2z}{y+4\frac{yz}{(x^3y^2z+y+z)}}}\ge 3\tag{J}$$

Now we introduce the inequality :

$$\sqrt{\frac{y}{z+\frac{y^{2}}{(xy^{2}z+y+z)}}}4\sqrt{\frac{y+2z}{y+4\frac{yz}{(xy^{2}z+y+z)}}}-3\geq 0$$

Or squaring :

$$f(x,y,z)=(16x^{2}y^{5}z^{2}+23x^{2}y^{4}z^{3}+23xy^{4}z+78xy^{3}z^{2}+10xy^{2}z^{3}+7y^{3}+10y^{2}z+26yz^{2}-13z^{3})/((xy^{2}z+y+5z)(xy^{2}z^{2}+y^{2}+yz+z^{2}))\ge 0$$

Now with a computer we have :

$$f(x,(y+z)/2,z)\geq 0$$

Now we can square $J$

Some idea to go further :

Replace in the first constraint :

$$\frac{uy-wy-4}{(u-w)(uy^2(w-v)+v^2y^2-vwy^2-1)}=z\tag{J}$$

Then use Am-Gm as we have :

$$Inequality to prove \geq 2\sqrt{\sqrt{\frac{y(u-w)}{z(u-w)+y^2(u-v)(v-w)}} \sqrt{\frac{y(v-w)+z(u-w)}{y(u-v)+y(v-w)z(u-w)}}}+\sqrt{\frac{z\left(u-w\right)+y\left(u-v\right)}{y\left(v-w\right)+z\left(u-w\right)y\left(u-v\right)}}$$

Now I think the inequality :

$$2\sqrt{\sqrt{\frac{y(u-w)}{z(u-w)+y^2(u-v)(v-w)}} \sqrt{\frac{y(v-w)+z(u-w)}{y(u-v)+y(v-w)z(u-w)}}}+\sqrt{\frac{z\left(u-w\right)+y\left(u-v\right)}{y\left(v-w\right)+z\left(u-w\right)y\left(u-v\right)}}\geq 3$$

Is true with $J$ then we can try to conclude using Young's inequality for product and a computer .

I stop here .