$\left(a^2+b^2\right)\left(c^2+d^2\right)=\left((a-b)^2+2ab\right)\left((c-d)^2+2cd\right)\ge 2\sqrt{(a-b)^2\cdot 2ab}\cdot 2\sqrt{(c-d)^2\cdot 2cd}=8\sqrt{abcd}|(a-b)(c-d)|$

Victoria_Discalceata1 wrote: $\left(a^2+b^2\right)\left(c^2+d^2\right)=\left((a-b)^2+2ab\right)\left((c-d)^2+2cd\right)\ge 2\sqrt{(a-b)^2\cdot 2ab}\cdot 2\sqrt{(c-d)^2\cdot 2cd}=8\sqrt{abcd}|(a-b)(c-d)|$ How did you come up with such a wonderful idea?

MS_Kekas wrote: For any positive real numbers $a, b, c, d$, prove the following inequality: $$(a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2) \geq 64abcd|(a-b)(b-c)(c-d)(d-a)|$$Proposed by Anton Trygub Generalization of ytChen : For any positive real numbers $a, b, c, d$, prove the following inequality:$$\prod_{i=1}^n\left(a_i^2+a_{i+1}^2\right) \ge \left(2\sqrt2\right)^n \prod_{i=1}^na_i\cdot\left| \prod_{i=1}^n(a_i-a_{i+1})\right|,$$where $a_{n+1}=a_1$.

sqing wrote: MS_Kekas wrote: For any positive real numbers $a, b, c, d$, prove the following inequality: $$(a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2) \geq 64abcd|(a-b)(b-c)(c-d)(d-a)|$$Proposed by Anton Trygub Generalization of ytChen : For any positive real numbers $a, b, c, d$, prove the following inequality:$$\prod_{i=1}^n\left(a_i^2+a_{i+1}^2\right) \ge \left(2\sqrt2\right)^n \prod_{i=1}^na_i\cdot\left| \prod_{i=1}^n(a_i-a_{i+1})\right|,$$where $a_{n+1}=a_1$. Note that $$a_i^2+a_{i+1}^2 = (a_i-a_{i+1})^2+2a_i a_{i+1} \stackrel{\text{AM-GM}}{\geq} 2\sqrt{2a_i a_{i+1}}\cdot\left|a_i-a_{i+1}\right|$$$$\implies \prod_{i=1}^n (a_i^2+a_{i+1}^2)\geq \prod_{i=1}^n 2\sqrt{2a_i a_{i+1}}\cdot\left|a_i-a_{i+1}\right| = \prod_{i=1}^n 2\sqrt2 \cdot \prod_{i=1}^n \sqrt{a_ia_{i+1}} \cdot \prod_{i=1}^n \left|a_i-a_{i+1}\right|= \left(2\sqrt2\right)^n \prod_{i=1}^n a_i \cdot \left|\prod_{i=1}^n (a_i-a_{i+1})\right|.\ \square$$