Let a, b, c and d be real numbers with a2+b2+c2+d2=4. Prove the inequality (a+2)(b+2)≥cd and give four numbers a, b, c and d such that equality holds.
Solution
The claimed inequality is equivalent to 2ab+4a+4b+8≥2cd, which can be written as 2ab+4a+4b+a2+b2+c2+d2+4≥2cd on account of the condition a2+b2+c2+d2=4. By the identity a2+b2+2ab+4a+4b+4=(a+b+2)2 we arrive at the equivalent and obvious inequality (a+b+2)2+(c−d)2≥0 The case of equality occurs for a+b=−2andc=d together with a2+b2+c2+d2=4. For instance when a=b=c=d=−1.
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Source: MathNet,
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