I solved the following inequality and I want to find others proof.
It would be great if you can share thoughts and ideas.
Let $a,b,c>0: ab+bc+ca=3$. Prove that: $$2(a+b+c)+3abc\ge bc\sqrt{4a^2+5}+ca\sqrt{4b^2+5}+ab\sqrt{4c^2+5}$$
My solution.
We can rewrite the inequality as $$\frac{2(a+b+c)}{abc}+3\ge \sqrt{4+\frac{5}{a^2}}+\sqrt{4+\frac{5}{b^2}}+\sqrt{4+\frac{5}{c^2}}$$ Set $p=a+b+c; q=ab+bc+ca=3; r=abc\le 1$. Using AM-GM: $$2\sqrt{4+\frac{5}{a^2}}\le \frac{9}{2a}+5-\frac{21}{2(a+2)}$$ Sum up similar isolated fudging, we obtain $$LHS\le \frac{27}{4r}-\frac{21(4p+15)}{4(r+4p+14)}+\frac{15}{2}$$ Thus, it suffices to prove: $$f(r)=(8p-27)(2p+7)+r(10p+18-9r)\ge 0$$ Which: $$f^{'}(r)=10p+18(1-r) >0$$ I divide into two cases:
$\bullet p\ge \dfrac{27}{8}$ : $f(r)\ge f(0)=(8p-27)(2p+7)\ge 0$
$\bullet 3\le p\le \dfrac{27}{8}$ : By Schur $$f(r)\ge f\left(\frac{12p-p^3}{9}\right)$$ $$=\frac{(p-3)\left((7-2p)(16p^4+104p^3+284p^2+1042p+1871)+5047\right)}{288}\ge 0$$ Thus, we are done. Equality holds iff $a=b=c=1$
Proof.
We may use AM-GM as \begin{align*} 2\sqrt{\frac{5}{a^2}+4}&=2\frac{1}{a}\cdot\sqrt{\frac{4a^2+5}{2a+\sqrt{bc}}\cdot(2a+\sqrt{bc})}\\&\le \frac{1}{a}\cdot\left(\frac{4a^2+5}{2a+\sqrt{bc}}+2a+\sqrt{bc}\right)\\&=\frac{1}{a}\cdot\left(\frac{4a^2+5}{2a+\sqrt{bc}}+\sqrt{bc}-2a+4a\right)\\&=\frac{1}{a}\cdot\left(\frac{bc+5}{2a+\sqrt{bc}}+4a\right)\\&\le \frac{1}{a}\cdot\left(\frac{bc+5}{2a+\dfrac{2bc}{b+c}}+4a\right)\\&=(b+c)\cdot \frac{bc+5}{6a}+4\\&=(a+b+c)\cdot \frac{bc+5}{6a}+4-\frac{bc+5}{6}. \end{align*}
It implies that $$(a+b+c)\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}+\frac{15}{abc}\right)+54\ge 12\sum_{cyc}\sqrt{\frac{5}{a^2}+4},$$ or $$(a+b+c)\left[(ab+bc+ca)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-2(a+b+c)+\frac{15}{abc}\right]+54\ge 12\sum_{cyc}\sqrt{\frac{5}{a^2}+4},$$ or $$\frac{3}{2}-\frac{(a+b+c)^2}{6}+\frac{2(a+b+c)}{abc}+3\ge\sum_{cyc}\sqrt{\frac{5}{a^2}+4}.$$ But $a+b+c\ge \sqrt{3(ab+bc+ca)}=3$ then $\dfrac{3}{2}-\dfrac{(a+b+c)^2}{6}\le 0.$
Henc