The only idea I have is to look for minimum of $f(x)=x^3\frac{1-b}{b}+2x^2d+x\left(\frac{b^2}{c}+\frac{c^2}{d}-4-d^2\right)+d^2$

ilove33 wrote: Let $a,b,c,d$ be positive real numbers satisfying $a+b+c+d=4$. Prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq 4+(a-d)^2$$ Proof: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}=a+b+c+d+\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-d)^2}{d}+\frac{(d-a)^2}{a}$$$$\geq a+b+c+d+\frac{(a-b+b-c+c-d)^2}{b+c+d}+\frac{(d-a)^2}{a}$$$$= a+b+c+d+(a-d)^2\left(\frac{1}{a}+\frac{1}{b+c+d}\right)\geq a+b+c+d+\frac{4(a-d)^2}{a+b+c+d}=4+(a-d)^2$$ sqing 2013: Let $a_1,a_2,\cdots,a_n$ be positive real numbers. Prove the inequality\[\frac{a_1^2}{a_2}+\frac{a_2^2}{a_3}+\cdots+

ilove33 wrote: Let $a,b,c,d$ be positive real numbers satisfying $a+b+c+d=4$. Prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq 4+(a-d)^2$$ We have \[\sum \frac{a^2}b - 4 = \sum \frac{(a-b)^2}b \geqslant \frac{(a-b+b-c+c-d+a-d)^2}{a+b+c+d} = (a-d)^2.\]