On Resolving a Fuss on Squares and Fractions over a few Inequalities

Question

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: $$\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2} + ...+\sqrt{a_n^2+b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2+(b_1+b_2+...+b_n)^2}$$

Answer 1

The second problem.

Let $a=\frac{x}{y}$ and $b=\frac{y}{z},$ where $x$, $y$ and $z$ are positives.

Thus, by C-S and Vasc we obtain: $$\sum_{cyx}\frac{1}{b(a+b)}=\sum_{cyc}\frac{1}{\frac{y}{z}\left(\frac{x}{y}+\frac{y}{z}\right)}=\sum_{cyc}\frac{z^2}{y^2+xz}=$$ $$=\sum_{cyc}\frac{z^4}{y^2z^2+xz^3}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x^2y^2+x^3y)}\geq\frac{(x^2+y^2+z^2)^2}{\frac{1}{3}(x^2+y^2+z^2)^2+\frac{1}{3}(x^2+y^2+z^2)^2}=\frac{3}{2}.$$ The Vasc's inequality it's the following. $$(x^2+y^2+z^2)^2\geq3(x^3y+y^3z+z^3x).$$ There are very many proofs of this beautiful inequality.

I think, the best of them it's the following: $$(x^2+y^2+z^2)^2-3(x^3y+y^3z+z^3x)=\frac{1}{2}\sum_{cyc}(x^2-y^2-xy-xz+2yz)^2\geq0.$$

Answer 2

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: $$\sum_{cyc}a^2b^2=(a^2+c^2)(b^2+d^2)\leq\left(\frac{a^2+c^2+b^2+d^2}{2}\right)^2=4.$$

Answer 3

The fourth problem it's just Minkowski (triangle inequality), but also, after squaring of the both sides, we can use C-S: $$\sqrt{(a_i^2+b_i^2)(a_j^2+b_j^2)}\geq a_ia_j+b_ib_j$$ and we obtain an identity.

Answer 4

The third problem.

By C-S and AM-GM twice we obtain: $$\sum_{cyc}\frac{d}{1+a^2b}=\sum_{cyc}\frac{d^2}{d+a^2bd}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(a+a^2bd)}=$$ $$=\frac{16}{4+ab(ad+bc)+cd(bc+ad)}=\frac{16}{4+(ab+cd)(ad+bc)}\geq$$ $$\geq\frac{16}{4+\left(\frac{ab+cd+ad+bc}{2}\right)^2}=\frac{16}{4+\left(\frac{(a+c)(b+d)}{2}\right)^2}\geq\frac{16}{4+\left(\frac{\left(\frac{a+c+b+d}{2}\right)^2}{2}\right)^2}=2.$$