$ab+bc+ca+abc=4$, prove $\sum\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}$

Question

Let $a,b,c> 0: ab+bc+ca+abc=4.$ Prove that: $$\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{2+\sqrt{bc}}{\sqrt{bc}+a}+\frac{2+\sqrt{ca}}{\sqrt{ca}+b}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}.$$

I try a well-known substitution $a=\dfrac{2x}{y+z}$ but I don't know how to continue because it is very complicated.

Also, if we set $a\rightarrow a^2,...$ and $a^2b^2+b^2c^2+c^2a^2+a^2b^2c^2=4,$ we will prove $$\sum_{cyc}\frac{ab+2}{ab+c^2}+\frac{a^4+b^4+c^4}{8a^2b^2c^2}\ge\frac{39}{8}$$ I think we can expand all and use $pqr$ method. So far, I am stuck here.

Please help me some better ideas. Thank you very much

Update

Is it possible to expanding all yields and how to end it ?

Answer 1

Proof.

We can use the isolated fudging idea.

Indeed, we will prove $$\frac{2+\sqrt{bc}}{\sqrt{bc}+a}\ge 2-\frac{2a}{a+b+c+bc}. \tag{1}$$ Taking cylic sum on $(1)$ it remains to prove that$$\frac{a^2+b^2+c^2+9abc}{16abc}\ge \sum_{cyc}\frac{a}{a+b+c+bc}, \tag{2}$$which is true by Cauchy-Schwarz.

Can you end it now ?

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Indeed, by CBS \begin{align*} \frac{2+\sqrt{bc}}{\sqrt{bc}+a}&=\frac{\sqrt{[a(b+c+bc)+bc][4a(b+c+bc)+(b+c+bc-a)^2]}+\sqrt{bc}(a+b+c+bc)}{\left(\sqrt{bc}+a\right)(a+b+c+bc)}\\&\ge \frac{2a(b+c+bc)+\sqrt{bc}(b+c+bc-a)+\sqrt{bc}(a+b+c+bc)}{\left(\sqrt{bc}+a\right)(a+b+c+bc)}\\&=\frac{2(b+c+bc)}{a+b+c+bc}. \end{align*} It implies$$\frac{2+\sqrt{bc}}{\sqrt{bc}+a}\ge 2-\frac{2a}{a+b+c+bc}. $$ Now, by using Cauchy - Schwarz$$\frac{a}{a+b+c+bc}=\frac{a}{3.\dfrac{a+b+c}{3}+bc}\le \frac{a}{16bc}+\frac{9a}{16(a+b+c)}.$$Similarly, we obtain $$\sum_{cyc}\frac{a}{a+b+c+bc}\le \sum_{cyc}\frac{a}{16bc}+\frac{9}{16}=\frac{a^2+b^2+c^2+9abc}{16abc}.$$ We end proof here. Equality holds at $a=b=c=1.$

Answer 2

Sketch of a proof.

We use the so-called isolated fudging.

It suffices to prove that $$\frac{2+\sqrt{ab}}{\sqrt{ab}+c} + \frac{a^2 + b^2}{16abc} \ge \frac{39}{8}\cdot \frac{6a + 6b + abc}{12(a + b + c) + 3abc}. \tag{1}$$ (Note: Take cyclic sum on (1), we get the desired inequality.)

(1) is true which is verified by Mathematica. I have a complicated proof below. I will try to find a better proof of (1) (e.g. by hand).

$\phantom{2}$

Let $p = a + b, q = ab$. Then $$p^2 \ge 4q, \quad 0 < q \le 4. \tag{2}$$ Using $c = \frac{4 - ab}{ab + a + b} = \frac{4 - q}{p + q}$, (1) is written as $$\frac{2+\sqrt{q}}{\sqrt{q} + \frac{4 - q}{p + q}} + \frac{p^2 - 2q}{16q \cdot \frac{4 - q}{p + q}} \ge \frac{39}{8}\cdot \frac{6p + q \cdot \frac{4 - q}{p + q}}{12p + 12 \cdot \frac{4 - q}{p + q} + 3q \cdot \frac{4 - q}{p + q}}. \tag{3}$$

It suffices to prove that (3) is true for all $0 < q \le 4$ and $p\ge \sqrt{4q}$.

Let $$u = \frac{2 - \sqrt{q}}{\sqrt{q}} \ge 0, \quad v = p - \sqrt{4q} \ge 0.$$ Then $$q = \frac{4}{(1 + u)^2}, \quad p = \frac{4}{1 + u} + v.$$

(3) is written as $$m_6 v^6 + m_5 v^5 + m_4 v^4 + m_3v^3 + m_2v^2 + m_1 v + m_0 \ge 0 \tag{4}$$ where \begin{align*} m_6 &= {u}^{10}+10\,{u}^{9}+45\,{u}^{8}+120\,{u}^{7}+210\,{u}^{6}+252\,{u}^{5 }\\ &\qquad +210\,{u}^{4}+120\,{u}^{3}+45\,{u}^{2}+10\,u+1,\\ m_5 &= 2\,{u}^{11}+22\,{u}^{10}+132\,{u}^{9}+540\,{u}^{8}+1548\,{u}^{7}+3108 \,{u}^{6}\\ &\qquad +4368\,{u}^{5}+4272\,{u}^{4}+2850\,{u}^{3}+1238\,{u}^{2}+316 \,u+36,\\ m_4 &= 44\,{u}^{10}+456\,{u}^{9}+2316\,{u}^{8}+7616\,{u}^{7}+17572\,{u}^{6}+ 28824\,{u}^{5}\\ &\qquad +33196\,{u}^{4}+26096\,{u}^{3}+13296\,{u}^{2}+3952\,u+ 520,\\ m_3 &= 8\,{u}^{11}+88\,{u}^{10}+1056\,{u}^{9}+6784\,{u}^{8}+24792\,{u}^{7}+ 59448\,{u}^{6}\\ &\qquad +102064\,{u}^{5}+128688\,{u}^{4}+115792\,{u}^{3}+69344\, {u}^{2}+24448\,u+3808,\\ m_2 &= 96\,{u}^{10}-256\,{u}^{9}-2208\,{u}^{8}+4960\,{u}^{7}+51200\,{u}^{6}+ 142048\,{u}^{5}\\ &\qquad +224512\,{u}^{4}+237760\,{u}^{3}+171072\,{u}^{2}+74752 \,u+14592,\\ m_1 &= 1344\,{u}^{9}+3392\,{u}^{8}-12736\,{u}^{7}-51136\,{u}^{6}-37760\,{u}^{ 5}+59520\,{u}^{4}\\ &\qquad +138752\,{u}^{3}+141312\,{u}^{2}+90112\,u+26624,\\ m_0 &= 4352\,{u}^{8}+19968\,{u}^{7}+15616\,{u}^{6}-34816\,{u}^{5}-33792\,{u}^ {4}\\ &\qquad +16384\,{u}^{3}-4096\,{u}^{2}+16384. \end{align*}

Clearly, $m_6, m_5, m_4, m_3 \ge 0$ (all coefficients are non-negative). It suffices to prove that $$m_2 v^2 + m_1 v + m_0 \ge 0. \tag{5}$$

It is easy to prove that $m_2, m_0 \ge 0$. By AM-GM, it suffices to prove that $$2\sqrt{m_2 m_0} + m_1 \ge 0$$ which is true.