Some Cauchy-Schwarz Inequalities

Question

I am trying to learn how to deal with inequalities to prepare for a Math Olympiad and right now I am working on Cauchy-Schwarz. However, I am not that good at seeing the relationships and I don't have the intuition to know how to apply the inequality. Right now I am stuck on a bunch of exercises so I am going to post them all here (all of them should be solved using C-S):
Ex 1:
Given $a^2+b^2+c^2+d^2=4$, a, b, c, and d are postive real numbers, prove $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge4$$ Ex 2: a, b, and c are positive real numbers, prove: $$(\frac{a}{a+2b})^2+(\frac{b}{b+2c})^2+(\frac{c}{c+2a})^2\ge\frac{1}{3}$$ Ex 3: a, b, c, and d are positive real numbers such that $a+b+c+d=4$, prove: $$\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\ge2$$ Ex 4: a, b, and c are positive real numbers such that $a+b+c=1$, prove: $$\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt2$$ I am asking a lot, but inequalities have always been my weak spot and I want to get over them. I would really appreciate some comments/explanations in the solutions, or some tips or good practices that I can in the future. Just 1 solution would be great. Thanks in advance!

Answer 1

1. 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 2

2. By C-S twice we obtain: $$\sum_{cyc}\left(\frac{a}{a+2b}\right)^2=\frac{1}{3}\sum_{cyc}1\sum_{cyc}\left(\frac{a}{a+2b}\right)^2\geq\frac{1}{3}\left(\sum_{cyc}\frac{a}{a+2b}\right)^2=$$ $$=\frac{1}{3}\left(\sum_{cyc}\frac{a^2}{a^2+2ab}\right)^2\geq\frac{1}{3}\left(\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+2ab)}\right)^2=\frac{1}{3}.$$

Answer 3

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

Answer 4

4. By C-S twice we obtain: $$\sum_{cyc}\frac{a}{\sqrt{a+2b}}=\sqrt{\left(\sum_{cyc}\frac{a}{\sqrt{a+2b}}\right)^2}\leq\sqrt{\sum_{cyc}a\sum_{cyc}\frac{a}{a+2b}}=$$ $$=\sqrt{3-\sum_{cyc}\left(1-\frac{a}{a+2b}\right)}=\sqrt{3-2\sum_{cyc}\frac{b}{a+2b}}=\sqrt{3-2\sum_{cyc}\frac{b^2}{ab+2b^2}}\leq$$ $$\leq\sqrt{3-\frac{2(a+b+c)^2}{\sum\limits_{cyc}(ab+2b^2)}}<\sqrt{3-\frac{2(a+b+c)^2}{2(a+b+c)^2}}=\sqrt2<2.$$