RockmanEX3 wrote: Let $a$, $b$, $c$ and $d$ be real numbers with $a^2 + b^2 + c^2 + d^2 = 4$. Prove that the inequality $$(a+2)(b+2) \ge cd$$holds and give four numbers $a$, $b$, $c$ and $d$ such that equality holds. (Walther Janous) This is from Russian mathematical olimpiad, 2015, grade 9: $$0 \leq (2 + a + b)^2 = 4 + 4(a + b) + (a + b)^2 = 8 + 4a + 4b + 2ab + a^2 + b^2 – 4 = 2(2 + a)(2 + b) – c^2 – d^2 \leq 2(2 + a)(2 + b) – 2cd.$$

Let $0<a,b,c,d<1$ and $abcd=(1-a)(1-b)(1-c)(1-d).$ Prove that $$a+b+c+d\geq 1+(a+c)(b+d).$$(ETS1331)

RockmanEX3 wrote: Let $a$, $b$, $c$ and $d$ be real numbers with $a^2 + b^2 + c^2 + d^2 = 4$. Prove that the inequality $$(a+2)(b+2) \ge cd$$holds and give four numbers $a$, $b$, $c$ and $d$ such that equality holds. (Walther Janous) Proof of Zhangyunhua: $$(a+2)(b+2) =\frac{1}{2}(c^2+d^2)+\frac{1}{2}(a+b+2)^2\ge cd.$$

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Wonderful problem and wonderful solution! Many thanks, professor Song Qing! Greetings!

Thank Lonesan.