Ex 1.11 [BMOC-2011]

[rakeen]

if $a+d=b+c$ then prove that $(a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c) \geq 0$







$a-b = c-d$ so, $(c-d)^2 + (b-d)^2 +(b-c)(d-a) \geq 0$
the inequality would be true if (b-c) and (d-a) are of same sign; meaning that $b\geq c$ and $d \geq a$. how can I prove it?

[rakeen]

$a-b = c-d$ so, $(c-d)^2 + (b-d)^2 +(b-c)(d-a) \geq 0$
the inequality would be true if (b-c) and (d-a) are of same sign; meaning that $b\geq c$ and $d \geq a$. how can I prove it?

[sm.joty]

You can try by this way. Try to prove,$(c-d)^{2}+(b-d)^{2}>(d-a)(b-c)$