Nice inequality with a,b,c,d

Question

If $a,b,c,d>0$ with $a+b+c+d=1$, prove that $\displaystyle \frac{bcd}{a+2}+\frac{acd}{b+2}+\frac{abd}{c+2}+\frac{abc}{d+2} <\frac{1}{13}$.

Answer 1

By AM-GM we obtain: $$\sum_{cyc}\frac{bcd}{a+2}=\frac{1}{2}\sum_{cyc}bcd+\sum_{cyc}\left(\frac{bcd}{a+2}-\frac{bcd}{2}\right)=$$ $$=\frac{1}{2}\sum_{cyc}bcd-\sum_{cyc}\frac{abcd}{2(a+2)}<\frac{1}{2}\sum_{cyc}bcd=$$ $$=\frac{abc+abd+acd+bcd}{2}=\frac{ab(c+d)+cd(a+b)}{2}\leq$$ $$\leq\frac{\left(\frac{a+b}{2}\right)^2(c+d)+\left(\frac{c+d}{2}\right)^2(a+b)}{2}=\frac{(a+b)(c+d)(a+b+c+d)}{8}= $$ $$=\frac{(a+b)(c+d)}{8}\leq\frac{\left(\frac{a+b+c+d}{2}\right)^2}{8}=\frac{1}{32}<\frac{1}{13}$$

Answer 2

Not so clear and "good" bound solution, but get necessary:

$(1)$ Use Cauchy inequality $bcd\leqslant (\tfrac{b+c+d}{3})^3$; $acd\leqslant (\tfrac{a+c+d}{3})^3$; $abd\leqslant (\tfrac{a+b+d}{3})^3$; $abc\leqslant (\tfrac{a+b+c}{3})^3$

as $a+b+c+d=1$, we get $bcd\leqslant (\tfrac{1-a}{3})^3$; $acd\leqslant (\tfrac{1-b}{3})^3$; $abd\leqslant (\tfrac{1-c}{3})^3$; $abc\leqslant (\tfrac{1-d}{3})^3$

So, $$\sum_{cyc}\frac{bcd}{a+2} \leqslant\tfrac{(1-a)^3}{27(a+2)}+\tfrac{(1-b)^3}{27(b+2)}+\tfrac{(1-c)^3}{27(c+2)}+\tfrac{(1-d)^3}{27(d+2)}$$

Then, as $0<a,b,c,d<1$ we can take value at zero in the right side inequality addends for further inequality estimation, because numerator increases and denominator decreases as $a,b,c,d$ approach to $0$.

Finally, we have: $$\sum_{cyc}\frac{bcd}{a+2} \leqslant\tfrac{(1-a)^3}{27(a+2)}+\tfrac{(1-b)^3}{27(b+2)}+\tfrac{(1-c)^3}{27(c+2)}+\tfrac{(1-d)^3}{27(d+2)}<\frac{1}{54}+\frac{1}{54}+\frac{1}{54}+\frac{1}{54}=\frac{1}{13.5}<\frac{1}{13}$$