<s>[INST] On Resolving a Fuss on Squares and Fractions over a few Inequalities Firstly, only AM-GM and C-S are to be sought.
$1.$Let $a, b, c, d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Show that
$${a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge 4$$
In my textbook, this problem is credited to Michael Rozenberg, but I couldn't find a solution to it by myself or on this site, so decided to ask.
I tried my best with fallacy-
$$4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2 \Rightarrow 4 \ge a+b+c+d$$ with the constraint given. Proceeding-
$$(a+b+c+d)\left({a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a}\right) \ge (a+b+c+d)^2 $$
$$\Rightarrow {a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge a+b+c+d$$
That means, I need $a+b+c+d\ge 4$ to complete the proof but instead got $4 \ge a+b+c+d$ !
$2.$Let $a, b, c$ be positive real numbers such that $abc = 1$.
Show that
$${1\over b(a+b)}+{1\over c(b+c)}+{1\over a(c+a)}\ge \frac{3}{2}.$$
$3.$If a, b, c and d are positive real numbers such that $a + b + c + d = 4$. Prove that
$$ {a \over 1+b^2c}+{b \over 1+c^2d}+{c \over 1+d^2a}+{d \over 1+a^2b} \ge 2. $$
The #2 is credited to the Zhautykov Olympiad 2008 and there is no need for revealing my attempts as I've no idea what to do, Lastly,
This is a doubt, not an question that definitely has an answer, but if the doubt comes out to have an answer, nevertheless, this is a problem.
$4.$ Prove without Induction:
$$\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2} + ...+\sqrt{a_n^2+b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2+(b_1+b_2+...+b_n)^2}$$
 [/INST] The first problem.
By Holder $$\left(\sum_{cyc}\frac{a^2}{b}\right)^2\sum_{cyc}a^2b^2\geq\left(\sum_{cyc}a^2\right)^3.$$
Thus, it's enough to prove that $$\sum_{cyc}a^2b^2\leq4,$$ which is true by AM-GM:
$$\sum_{cyc}a^2b^2=(a^2+c^2)(b^2+d^2)\leq\left(\frac{a^2+c^2+b^2+d^2}{2}\right)^2=4.$$
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