If $a+b+c+d=4$ Prove that $ \sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4 $ Question -
Let $a, b, c, d$ be non-negative real numbers with sum 4. Prove that
$
\sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4
$
My work -
first i multiply both numerator and denominator by $\sqrt{ab+1}$ and i apply CS in numerator but in end it does not work..
now i multiply both numerator and denominator by $\sqrt{a+1}$ and apply holder but it also fails..
i also try some substitutions but none of them work
In solution to this problem author apply am-gm and we need to prove 
$(a+1)(b+1)(c+1)(d+1) \geq(a b+1)(b c+1)(c d+1)(d a+1)$
and he proves it by expanding , i understand his proof but can someone solve this problem using classic inequalities without using such a boring expansion ???

Yes, your second inequality is also true.
Indeed, after expanding we need to prove that:
$$a+b+c+d+ac+bd+\sum_{cyc}abc\geq\sum_{cyc}a^2bd+abcd\sum_{cyc}ab+a^2b^2c^2d^2+abcd.$$
Now, by AM-GM $$4=a+b+c+d\geq4\sqrt[4]{abcd},$$ which gives $$abcd\leq1.$$
Thus, by AM-GM again we obtain:
$$ac+bd\geq2\sqrt{abcd}\geq abcd+a^2b^2c^2d^2.$$
Also, by AM-GM again $$\sum_{cyc}ab=(a+c)(b+d)\leq\left(\frac{a+c+b+d}{2}\right)^2=4.$$
Thus, by AM-GM again:
$$\sum_{cyc}abc\geq4\sqrt[4]{a^3b^3c^3d^3}\geq 4abcd\geq abcd\sum_{cyc}ab.$$
Id est, it's enough to prove that
$$\sum_{cyc}a^2bd\leq4,$$ which is true by  AM-GM twice:
$$\sum_{cyc}a^2bd=(ab+cd)(ad+bc)\leq\left(\frac{ab+bc+cd+da}{2}\right)^2\leq4.$$
Done!
We can write this solution in one line:
$$\prod_{cyc}(1+ab)=(1+ab)(1+cd)(1+bc)(1+da)=$$
$$=(1+abcd+ab+cd)(1+abcd+bc+da)=$$
$$=(1+abcd)^2+(1+abcd)(ab+bc+cd+da)+(ab+cd)(bc+da)\leq$$
$$\leq1+2abcd+a^2b^2c^2d^2+\sum_{cyc}ab+abcd\sum_{cyc}ab+4=$$
$$=5+abcd+(abcd+a^2b^2c^2d^2)+\sum_{cyc}ab+abcd\sum_{cyc}ab\leq$$
$$\leq5+abcd+ac+bd+\sum_{cyc}ab+\sum_{cyc}abc=\prod_{cyc}(1+a).$$
