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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 $\Leftrightarrow a=b=c=d=\frac{1}{2}.$Bài toán xong !!!Đúng thì vote up, sai thì đừng báo cáo vi phạm nha :D

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 !!!

$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 $\Leftrightarrow a=b=c=d=\frac{1}{2}.$Bài toán xong !!!Đúng thì vote up, sai thì đừng báo cáo vi phạm nha :D

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 $\Leftrightarrow a=b=c=d=\frac{1}{2}.$Bài toán xong !!!Đúng thì vote up, sai thì đừng báo cáo vi phạm nha :D