Prove that: $\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}+\frac{1}{d+3}\leq1$ Let $a$, $b$, $c$ and $d$ are non-negative numbers such that $abc+abd+acd+bcd=4.$ Prove that:
$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}\leq1$
I simplified it and it turns out that it suffices to show
$19\le3(ab+ac+ad+bc+bd+cd)+abcd$
subject to $abc+abd+acd+bcd=4.$
I then tried Lagrange multiplier but it didn't help too much...

Let $a+b+c+d=3u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Hence, we need to prove that
$$\prod_{cyc}(a+3)\geq\sum_{cyc}(a+3)(b+3)(c+3)$$ or
$$3(ab+ac+bc+ad+bd+cd)+2(abc+abd+acd+bcd)+abcd\geq27$$ or
$$18v^2w^2+8w^4+t^4\geq27w^4$$ or
$$19w^4-18v^2w^2-t^4\leq0$$ or
$$w^2\leq\frac{9v^2+\sqrt{81v^4+19t^4}}{19}.$$
But $$24(3v^4-4uw^3+t^4)=(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2\geq0,$$
which gives $t^4\geq4uw^3-3v^4$.
Hence, it remains to prove that
$$w^2\leq\frac{9v^2+\sqrt{81v^4+19(4uw^3-3v^4)}}{19}$$ or
$$9v^2+\sqrt{24v^6+76uw^3}\geq19w^2,$$
which is obvious by Maclaurin: $u\geq w$ and $v^2\geq w^2$.
A proof of Maclaurin see here:  How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$
Done!
