Do $a+b+c+d=2$ nên luôn có ít nhất $1$ số $>0$.
$TH1:$Với $a,b,c,d>0$. Ta có:
$VT=4-\sum_{}^{}\frac{3a^2}{1+3a^2}=4-\sum_{}^{}\frac{3a^2}{1+4a^2-a^2}\geq 4-\sum_{}^{}\frac{3a^2}{4a-a^2} $
$=4+\sum_{}^{}\frac{3a}{a-4}.$ $(1)$
Có: $\sum_{}^{}\frac{3a}{a-4}=\sum_{}^{} \frac{3(a-4)}{a-4}+\sum_{}^{}\frac{12}{a-4}=12+12.\sum_{}^{}\frac{1}{a-4} $
$\geq 12+12.\frac{(1+1+1+1)^2}{a+b+c+d-16}=-\frac{12}{7}.$
Thay vào $(1)$ $\Rightarrow VT\geq 4-\frac{12}{7}=\frac{16}{7}.$
Dấu $=$ xảy ra $\Leftrightarrow a=b=c=d=\frac{1}{2}.$
$TH2:$Với $a,b,c>0;d=0$
$VT=1+\sum_{}^{}\frac{1}{1+3a^2}=4-\sum_{}^{}\frac{3a^2}{3a^2+1}=4-\sum_{}^{}\frac{3a^2}{\frac{1}{3}(9a^2+4)-\frac{1}{3}} $
$\geq 4+3.\sum_{}^{}\frac{a^2}{\frac{1}{3}-4a}\geq 4+3.\frac{(a+b+c)^2}{1-4(a+b+c)}=\frac{16}{7}$.
Dấu $=$ xảy ra khi $d=0;a=b=c=\frac{2}{3}.$
$TH3:$
$a,b>0;c=d=0$.
$VT=2+\sum_{}^{}\frac{1}{1+3a^2}=4-\sum_{}^{} \frac{3a^2}{3(a^2+1)-2}\geq 4+\sum_{}^{}\frac{3a^2}{2-6a}\geq 4+3.\sum_{}^{}\frac{(a+b)^2}{4-6(a+b)} =\frac{5}{2}>\frac{16}{7}.$
$TH4:a=2,b=c=d=0$.Bất đẳng thức thỏa mãn.
Bài toán xong !!!