Equality holds when $a=b=c=d$. Since both equality and inequality are homogenous, we can assume that $d=1$. \[\frac{(a+b+c)^2}{2(a+b+c)+3}\leq \sum{\frac{a^2}{b+c+1}=\frac{3}{a+b+c}}\implies (a+b+c-3)((a+b+c)^2+3(a+b+c)+3)\leq 0\]Hence $a+b+c\leq 3$. \[\sum{\frac{16}{b+c+2}}\leq \sum{(\frac{4}{b+1}+\frac{4}{c+1})}=\sum{\frac{8}{a+1}}\]\[\sum{\frac{a+3}{a}}\overset{?}{\geq}\sum{\frac{8}{a+1}}\iff \sum{\frac{(1-a)(3-a)}{a(a+1)}}\overset{?}{\geq}0\]WLOG $a\geq b\geq c$. Note that $\frac{1}{a}-1\leq \frac{1}{b}-1\leq \frac{1}{c}-1$ and $\frac{4}{a+1}-1\leq \frac{4}{b+1}-1\leq \frac{4}{c+1}-1$. By Chebyshev, we have \[\sum{(\frac{1}{a}-1)(\frac{4}{a+1}-1)}\geq (\sum{\frac{1}{a}}-3)(\sum{\frac{4}{a+1}}-3)\geq 0\]Because $\sum{\frac{1}{a}}\geq \frac{9}{a+b+c}\geq 3$ and $\sum{\frac{4}{a+1}}\geq \frac{36}{a+b+c+3}>3$ as desired. $\blacksquare$

bin_sherlo wrote: Equality holds when $a=b=c=d$. Since both equality and inequality are homogenous, we can assume that $d=1$. \[\frac{(a+b+c)^2}{2(a+b+c)+3}\leq \sum{\frac{a^2}{b+c+1}=\frac{3}{a+b+c}}\implies (a+b+c-3)((a+b+c)^2+3(a+b+c)+3)\leq 0\]Hence $a+b+c\leq 3$. \[\sum{\frac{16}{b+c+2}}\leq \sum{(\frac{4}{b+1}+\frac{4}{c+1})}=\sum{\frac{8}{a+1}}\]\[\sum{\frac{a+3}{a}}\overset{?}{\geq}\sum{\frac{8}{a+1}}\iff \sum{\frac{(1-a)(3-a)}{a(a+1)}}\overset{?}{\geq}0\]WLOG $a\geq b\geq c$. Note that $\frac{1}{a}-1\leq \frac{1}{b}-1\leq \frac{1}{c}-1$ and $\frac{4}{a+1}-1\leq \frac{4}{b+1}-1\leq \frac{4}{c+1}-1$. By Chebyshev, we have \[\sum{(\frac{1}{a}-1)(\frac{4}{a+1}-1)}\geq (\sum{\frac{1}{a}}-3)(\sum{\frac{4}{a+1}}-3)\geq 0\]Because $\sum{\frac{1}{a}}\geq \frac{9}{a+b+c}\geq 3$ and $\sum{\frac{4}{a+1}}\geq \frac{36}{a+b+c+3}>3$ as desired. $\blacksquare$ How can you assume $d=1$

m4thbl3nd3r wrote: bin_sherlo wrote: Equality holds when $a=b=c=d$. Since both equality and inequality are homogenous, we can assume that $d=1$. \[\frac{(a+b+c)^2}{2(a+b+c)+3}\leq \sum{\frac{a^2}{b+c+1}=\frac{3}{a+b+c}}\implies (a+b+c-3)((a+b+c)^2+3(a+b+c)+3)\leq 0\]Hence $a+b+c\leq 3$. \[\sum{\frac{16}{b+c+2}}\leq \sum{(\frac{4}{b+1}+\frac{4}{c+1})}=\sum{\frac{8}{a+1}}\]\[\sum{\frac{a+3}{a}}\overset{?}{\geq}\sum{\frac{8}{a+1}}\iff \sum{\frac{(1-a)(3-a)}{a(a+1)}}\overset{?}{\geq}0\]WLOG $a\geq b\geq c$. Note that $\frac{1}{a}-1\leq \frac{1}{b}-1\leq \frac{1}{c}-1$ and $\frac{4}{a+1}-1\leq \frac{4}{b+1}-1\leq \frac{4}{c+1}-1$. By Chebyshev, we have \[\sum{(\frac{1}{a}-1)(\frac{4}{a+1}-1)}\geq (\sum{\frac{1}{a}}-3)(\sum{\frac{4}{a+1}}-3)\geq 0\]Because $\sum{\frac{1}{a}}\geq \frac{9}{a+b+c}\geq 3$ and $\sum{\frac{4}{a+1}}\geq \frac{36}{a+b+c+3}>3$ as desired. $\blacksquare$ How can you assume $d=1$ Degree of both sides are $1$ in equality and degree of both sides are $-1$ in inequality thus we can assume $d=1$. When you multiply $(a,b,c,d)$ by $\frac{1}{d},$ the condition and inequality don't get changed.

$\frac{(a+b+c)^2}{2(a+b+c)+3d} \leq \dfrac{a^2}{b+c+d} + \dfrac{b^2}{a+c+d} + \dfrac{c^2}{a+b+d} = \dfrac{3d^2}{a+b+c} \to a+b+c \leq 3d$ $\dfrac{3}{a}+ \dfrac{3}{b} + \dfrac{3}{c}+ \dfrac{3}{d} \geq \dfrac{2}{a}+ \dfrac{2}{b} + \dfrac{2}{c}+ \frac{9}{a+b+c} +\dfrac{3}{d} \geq \dfrac{2}{a}+ \dfrac{2}{b} + \dfrac{2}{c}+ \dfrac{6}{d} \geq \dfrac{16}{a+b+2d} + \dfrac{16}{b+c+2d} + \dfrac{16}{a+c+2d}$ because $\frac{1}{a}+\frac{1}{b}+\frac{2}{d} \geq \frac{16}{a+b+2d}$

By Radon's Inequality $$\dfrac{a^2}{b+c+d}+\dfrac{b^2}{c+d+a}+\dfrac{c^2}{d+a+b}=\dfrac{3d^2}{a+b+c}\geq \dfrac{(a+b+c)^2}{2(a+b+c)+3d}$$which implies $a+b+c\leq 3d$. From now on use Radon's Inequality $$\dfrac{16}{a+b+2d}\leq \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{d}$$Thus, $$RHS\leq 2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{6}{d}\overbrace{\leq}^{?} 3\sum_{cyc}{\dfrac{1}{a}}$$$$\Longleftrightarrow \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq \dfrac{3}{d}\Longleftrightarrow a+b+c\leq 3d$$as desired.

bin_sherlo wrote: m4thbl3nd3r wrote: bin_sherlo wrote: Equality holds when $a=b=c=d$. Since both equality and inequality are homogenous, we can assume that $d=1$. \[\frac{(a+b+c)^2}{2(a+b+c)+3}\leq \sum{\frac{a^2}{b+c+1}=\frac{3}{a+b+c}}\implies (a+b+c-3)((a+b+c)^2+3(a+b+c)+3)\leq 0\]Hence $a+b+c\leq 3$. \[\sum{\frac{16}{b+c+2}}\leq \sum{(\frac{4}{b+1}+\frac{4}{c+1})}=\sum{\frac{8}{a+1}}\]\[\sum{\frac{a+3}{a}}\overset{?}{\geq}\sum{\frac{8}{a+1}}\iff \sum{\frac{(1-a)(3-a)}{a(a+1)}}\overset{?}{\geq}0\]WLOG $a\geq b\geq c$. Note that $\frac{1}{a}-1\leq \frac{1}{b}-1\leq \frac{1}{c}-1$ and $\frac{4}{a+1}-1\leq \frac{4}{b+1}-1\leq \frac{4}{c+1}-1$. By Chebyshev, we have \[\sum{(\frac{1}{a}-1)(\frac{4}{a+1}-1)}\geq (\sum{\frac{1}{a}}-3)(\sum{\frac{4}{a+1}}-3)\geq 0\]Because $\sum{\frac{1}{a}}\geq \frac{9}{a+b+c}\geq 3$ and $\sum{\frac{4}{a+1}}\geq \frac{36}{a+b+c+3}>3$ as desired. $\blacksquare$ How can you assume $d=1$ Degree of both sides are $1$ in equality and degree of both sides are $-1$ in inequality thus we can assume $d=1$. When you multiply $(a,b,c,d)$ by $\frac{1}{d},$ the condition and inequality don't get changed. Thanks

bin_sherlo wrote: Equality holds when $a=b=c=d$. Since both equality and inequality are homogenous, we can assume that $d=1$. \[\frac{(a+b+c)^2}{2(a+b+c)+3}\leq \sum{\frac{a^2}{b+c+1}=\frac{3}{a+b+c}}\implies (a+b+c-3)((a+b+c)^2+3(a+b+c)+3)\leq 0\]Hence $a+b+c\leq 3$. \[\sum{\frac{16}{b+c+2}}\leq \sum{(\frac{4}{b+1}+\frac{4}{c+1})}=\sum{\frac{8}{a+1}}\]\[\sum{\frac{a+3}{a}}\overset{?}{\geq}\sum{\frac{8}{a+1}}\iff \sum{\frac{(1-a)(3-a)}{a(a+1)}}\overset{?}{\geq}0\]WLOG $a\geq b\geq c$. Note that $\frac{1}{a}-1\leq \frac{1}{b}-1\leq \frac{1}{c}-1$ and $\frac{4}{a+1}-1\leq \frac{4}{b+1}-1\leq \frac{4}{c+1}-1$. By Chebyshev, we have \[\sum{(\frac{1}{a}-1)(\frac{4}{a+1}-1)}\geq (\sum{\frac{1}{a}}-3)(\sum{\frac{4}{a+1}}-3)\geq 0\]Because $\sum{\frac{1}{a}}\geq \frac{9}{a+b+c}\geq 3$ and $\sum{\frac{4}{a+1}}\geq \frac{36}{a+b+c+3}>3$ as desired. $\blacksquare$ After obtaining \[\sum{\frac{16}{b+c+2}}\leq \sum{(\frac{4}{b+1}+\frac{4}{c+1})}=\sum{\frac{8}{a+1}}\]we can also use the inequality \[a \geq \dfrac{8}{a+1}-\dfrac{3}{a} \quad \forall a \in \mathbb{R}^{+} \]which one shots the problem.