Quote: Let $a,b,c$ be positive real numbers satisfying $a^2+b^2+c^2 \geq 3.$ Prove that \[ \frac{(a+1)(b+2)}{(b+1)(b+5)} + \frac{(b+1)(c+2)}{(c+1)(c+5)}+\frac{(c+1)(a+2)}{(a+1)(a+5)} \geq \frac{3}{2} \]

I want to give a motivation for the solution of this problem. The key is to come up with a polynomial (of $b+2$), so as to bound $(b+1)(b+5)$. For doing so, one way is to consider, $t=b+2$, then, $(b+1)(b+5)=(t-1)(t+3)=t^2+2t-3\leq ct^2$, where $c>0$ is the constant to be determined. We want to preserve an equality case (which happens at $a=b=c=1$, in particular, for $t=3$), giving immediately, $c=\frac{4}{3}$. After doing so, the rest is just proving $\sum_{cyc}\frac{a+1}{b+2}\geq 2$, and a Cauchy-Schwarz (with weights $(a+1)(b+2)$, $(b+1)(c+2)$, and, $(c+1)(a+2)$) will do the work.

how to prove this $\frac{((\sum a) +3)^2}{\sum ab + 3 \sum a +6} \geq 2$

Equality holds when $a=b=c=1$. Let $f(x)=\frac{(x+2)}{(x+1)(x+5)}$ Claim: $f$ is convex. Proof: $f'(x)=\frac{-(x^2+4x+7)}{(x^2+6x+5)^2}$ $f''(x)=\frac{(-2x-4)((x^2+6x+5)^2-(4x^3+36x^2+92x+60)(-(x^2+4x+7))}{(x^2+6x+5)^4}=\frac{(x^2+4x+7)(4x^3+36x^2+92x+60)-(2x+4)(x^2+6x+5)^2}{(x^2+6x+5)^4}$ $(x^2+4x+7)(4x^3+36x^2+92x+60)-(2x+4)(x^4+12x^3+46x^2+60x+25)=(4x^5+36x^4+92x^3+60x^2+16x^4+368x^2+240x+28x^3+252x^2+574x+420)-(2x^5+24x^4+82x^3+120x^2+50x+4x^4+48x^3+184x^2+240x+100)$ $=4x^5+52x^4+120x^3+680x^2+814x+420-2x^5-28x^4-130x^3-304x^2-290x-100=2x^5+24x^4-10x^3+376x^2+524x+320>0$ By Jensenn, we have \[\frac{LHS}{a+b+c+3}=\frac{(a+1)}{a+b+c+3}.f(b)+\frac{(b+1)}{a+b+c+3}.f(c)+\frac{(c+1)}{a+b+c+3}.f(a)\geq f(\frac{ab+bc+ca+a+b+c}{a+b+c+3})\]Let's write $a+b+c=3u, ab+bc+ca=3v^2$ We have $9u^2-6v^2\geq 3\iff 3u^2-2v^2\geq 1$ \[f(\frac{ab+bc+ca+a+b+c}{a+b+c+3})=f(\frac{u+v^2}{u+1})=\frac{(\frac{v^2+3u+2}{u+1})}{(\frac{v^2+2u+1}{u+1})(\frac{v^2+6u+5}{u+1})}=(u+1)\frac{(v^2+3u+2)}{(v^2+2u+1)(v^2+6u+5)}\]\[(u+1)\frac{(v^2+3u+2)}{(v^2+2u+1)(v^2+6u+5)}\overset{?}\geq \frac{1}{2(u+1)}\]\[2(u+1)^2\overset{?}\geq \frac{(v^2+2u+1)(v^2+6u+5)}{v^2+3u+2}=(v^2+2u+1)\frac{(v^2+6u+5)}{(v^2+3u+2)}=(v^2+2u+1)(1+\frac{3u+3}{v^2+3u+2})\]\[2(u+1)^2-(3-\frac{3u+3}{v^2+3u+2})(u+1)-v^2-2u-1 \overset{?}\geq 0\]\[2(u+1)^2-3(u+1)+\frac{3(u+1)^2}{v^2+3u+2}-(v^2+2u+1)\overset{?}\geq 0\]\[(u+1)^2(2+\frac{3}{v^2+3u+2})\overset{?}\geq v^2+5u+4\]We have $v^2\leq \frac{3u^2-1}{2}$ so \[LHS\geq (u+1)^2(2+\frac{3}{\frac{3u^2-1}{2}+3u+2})=2(u+1)^2+\frac{6(u+1)^2}{3u^2-6u+3}=2(u+1)^2+\frac{6(u+1)^2}{3(u-1)^2}=2u^2+4u+4\]\[RHS=v^2+5u+4\leq \frac{3u^2-1}{2}+5u+4=\frac{3u^2+10u+7}{2}\]Also \[2u^2+4u+4\geq \frac{3u^2+10u+7}{2}\iff u^2-2u+1\geq0\iff (u-1)^2\geq 0\]which is true.$\blacksquare$