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 ?
Best Answer
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 ?
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 $a=b=c=1.$