Question

Demonstrate that: a2+b2+c2+d2 ≥ (a +b)(c+d)
(a#0)

Answer 1

Hi,

Explanation:

`To\ prove\ that \\a^2+b^2+c^2+d^2\geq (a+b)(c+d)\\where\ a\\eq 0\\Expanding\ the\ right-hand side,\ we\ have:\ (a+b)(c+d)=ac+ad+bc+bd\\`

`(a-c)^2\geq 0\\(b-d)^2\geq 0\\(a-c)^2+(b-d)^2\geq 0\\a^2-2ac+c^2+b^2-2bd+d^2\geq 0\\\\a^2+c^2+b^2+d^2\geq 2ac+2bd\ (1)\\`

`(a-d)^2\geq 0\\(b-c)^2\geq 0\\(a-d)^2+(b-c)^2\geq 0\\a^2+d^2-2ad+b^2+c^2-2bc\geq 0\\\\a^2+c^2+b^2+d^2\geq 2ad+2bc\ (2)\\`

Adding these two inequalities, we get:

`2(a^2+b^2+c^2+d^2)\geq 2ac+2bd+2ad+2bc\\\\Dividing\ both\ sides\ by\ 2, we\ obtain:\\a^2+b^2+c^2+d^2\geq ac+bd+ad+bc\\\\Simplifying\ further, we\ have:\\a^2+b^2+c^2+d^2\geq (a+b)(c+d)\\\\`