Problem: Let $a,b,c,d>0.$ Prove that: $$\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt{3}\ge2\left(\sqrt{\frac{d}{a+b+c}}+\frac{\sqrt{ab+bc+ca}}{a+b+c}\right)$$ It is my old teacher's problem.
My attempt: I guess a=b=c=d so using C-S inequality: $$R.H.S\le2\sqrt{\frac{d}{a+b+c}+\frac{ab+bc+ca}{(a+b+c)^2}}$$ But it seems weak to get the proof. I hope we can find a good solution for nice problem.
Thank you!
We need to prove that: $$\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}\geq2\sqrt{\frac{d}{a+b+c}}$$ and since $$\sqrt3(a+b+c)\geq2\sqrt{ab+ac+bc},$$ by AM-GM we obtain: $$\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}\geq2\sqrt{\frac{d\left(\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}\right)}{\sqrt{a^2+b^2+c^2}}}.$$ Thus, it's enough to prove that: $$\frac{\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}}{\sqrt{a^2+b^2+c^2}}\geq\frac{1}{a+b+c}.$$ Now, let $a^2+b^2+c^2=t(ab+ac+bc).$
Thus, we need to prove that: $$\frac{\sqrt3-\frac{2}{\sqrt{t+2}}}{\sqrt{t}}\geq\frac{1}{\sqrt{t+2}}$$ or $$\sqrt{3(t+2)}\geq2+\sqrt{t},$$ which is true by C-S: $$\sqrt{3(t+2)}=\sqrt{(2+1)(2+t)}\geq2+\sqrt{t}.$$