A friend of mine communicated me to ask a problem. I tried to prove and saw that it is not simple.
Let $a,b,c\ge 0: ab+bc+ca=3.$ Find maximum$$T=\frac{\sqrt{5ab+4}+\sqrt{5bc+4}+\sqrt{5ca+4}}{\sqrt{a+b+c+abc}}.$$
My prediction: $T $ achieve maximal value $\dfrac{9}{2}$ when $a=b=c=1.$
I do with symmetric inequalities before and equality almost happens at all equal variables.
If it is not cases for this problem, I'd like to know the reason. I would very much appreciate comments and additional examples.
I checked $a=b=\sqrt{3};c=0.$ The value $\dfrac{9}{2}$ is the best, I think.
Thus, we need to prove$$\sqrt{5ab+4}+\sqrt{5bc+4}+\sqrt{5ca+4}\le \frac{9}{2}\sqrt{a+b+c+abc}. \tag{1}$$ I can easily prove $$\sqrt{5ab+4}+\sqrt{5bc+4}+\sqrt{5ca+4}\le 3\sqrt{3(a+b+c)}.$$ By using CBS inequality $$\sqrt{5ab+4}+\sqrt{5bc+4}+\sqrt{5ca+4}\le \sqrt{3}\sqrt{12+5(ab+bc+ca)}=9\le 3\sqrt{3(a+b+c)}.$$ But when I use CBS for $(1)$ $$\sqrt{5ab+4}+\sqrt{5bc+4}+\sqrt{5ca+4}\le 9\le \frac{9}{2}\sqrt{a+b+c+abc}$$where the last one is equivalent to $a+b+c+abc\ge 4.$ It is trivially wrong at $a=b=\sqrt{3};c=0.$
Thus, how can I prove it in correct way? Hope you help me out. Thank you.