Looking for a tricky proof for two equality occuring $ab+bc+ca=3.$

Question

I'm looking a non trivial proof for the following inequality. It might be comlicated but easily verified by hand.

Let $a,b,c\ge 0: ab+bc+ca=3.$ Prove that $$\sqrt{ab+1}+\sqrt{bc+1}+\sqrt{ca+1}\ge \sqrt{2abc+16}.$$ Equality occurs when $(a,b,c)=(1,1,1); (0,\sqrt{3},\sqrt{3}).$

My strategy is proving a radical through non-radical inequalities.

Let $$x=\sqrt{\frac{ab+1}{2abc+16}};y=\sqrt{\frac{bc+1}{2abc+16}};z=\sqrt{\frac{ca+1}{2abc+16}}.$$We need to prove $$x+y+z\ge 1.$$Possible values for holding equality are $x=y=z=1/3$ and $x=y=1/4,z=1/2.$

The main idea is prove an appropriate fact about relations between $x^2y^2z^2,x^2+y^2+z^2,x^2y^2+y^2z^2+z^2x^2,...$

We can see also an answer of my topic Topic 1, Topic 2.

As @RiverLi wrote remark, "Then we seek some relations among $x^2,y^2,z^2$ which can implies $f(x,y,z)\ge 0$. The relations among $x^2,y^2,z^2$ clearly contain no radicals, sometimes it is easy to handle."

Also, this trick gives a short and nice proof. I found similar topic Topic 3.

Apparently, it is not easy and works in some inequalities. Back on the main problem, I test the trick with try and error without success.

Could someone who is interest in tricky proof give some hints to use this idea effectively?

Is there a systematic approach to motivate your approach?

Hope to see some useful answers. Also, all ideas about solutions are welcome.

Answer 1

Let $x = ab, y = bc, z = ca$. Then $abc = \sqrt{xyz}$.

It suffices to prove that, for all $x, y, z \ge 0$ with $x + y + z = 3$, $$\sqrt{x + 1} + \sqrt{y + 1} + \sqrt{z + 1} \ge \sqrt{2\sqrt{xyz} + 16}. \tag{1}$$

Let $$u = \sqrt{x + 1} - 1, \quad v = \sqrt{y + 1} - 1, \quad w = \sqrt{z + 1} - 1.$$ We have $u, v, w \ge 0$ and $x = u^2 + 2u, y = v^2 + 2v, z = w^2 + 2w$.

The condition $x + y + z = 3$ is written as $$u^2 + 2u + v^2 + 2v + w^2 + 2w = 3. \tag{2}$$

From (2), we have $$1 \le u + v + w \le 3\sqrt{2} - 3. \tag{3}$$ (Note: Using $u^2 + v^2 + w^2 \le (u + v + w)^2$ and (2), we have $(u + v + w)^2 + 2(u + v + w) \ge 3$ which gives $u + v + w \ge 1$. Using $u^2 + v^2 + w^2 \ge \frac{(u + v + w)^2}{3}$ and (2), we have $\frac{(u + v + w)^2}{3} + 2(u + v + w) \le 3$ which gives $u + v + w \le 3\sqrt{2} - 3$. )

It suffices to prove that $$u + v + w + 3 \ge \sqrt{2\sqrt{(u^2 + 2u)(v^2 + 2v)(w^2 + 2w)} + 16}$$ or $$(u + v + w + 3)^2 - 16 \ge 2\sqrt{(u^2 + 2u)(v^2 + 2v)(w^2 + 2w)}$$ or (using (3)) $$[(u + v + w + 3)^2 - 16]^2 \ge 4(u^2 + 2u)(v^2 + 2v)(w^2 + 2w). \tag{4}$$

We use the pqr method.

Let $p = u + v + w, q = uv + vw + wu, r = uvw$. From (3), we have $1\le p \le 3\sqrt{2} - 3$.

The condition (2) is written as $p^2 - 2q + 2p = 3$ which results in $q = \frac{p^2 + 2p - 3}{2}$. Using $q^2 \ge 3pr$, we have $$r \le \frac{q^2}{3p} = \frac{(p^2 + 2p - 3)^2}{12p}. \tag{5}$$

(4) is written as $$[(p + 3)^2 - 16]^2 \ge 4r^2+4(p^2+6p+5)r. \tag{6}$$

From (5) and (6), it suffices to prove that $$[(p + 3)^2 - 16]^2 \ge 4\cdot \left(\frac{(p^2 + 2p - 3)^2}{12p}\right)^2+4(p^2+6p+5)\cdot \frac{(p^2 + 2p - 3)^2}{12p} $$ or $$\frac{(9 - 6p - p^2)(p^3+17p^2+69p+9)(p-1)^3}{36p^2} \ge 0$$ which is true for all $1\le p\le 3\sqrt{2} - 3$.

We are done.