Question

let a,b,c,d be real numbers such that a+d=b+c.prove that (a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c)≥0 holds true​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{a+d=b+c}$

$\sf{\implies\,a-b=c-d\,\,\,\,\,\,...(1)}$

Again,

$\sf{\implies\,a-c=b-d\,\,\,\,\,\,\,...(2)}$

Also,

$\sf{\implies\,a+d-a+a=b+c}$

$\sf{\implies\,d-a=b+c-2a\,\,\,\,\,\,\,\,...(3)}$

Now,

$\tt{(a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c)}$

From (1), (2), and (3), we get,

$\tt{=(a-b)(a-b)+(a-c)(a-c)+(b+c-2a)(b-c)}$

$\tt{=(a-b)^2+(a-c)^2+(b+c)(b-c)-2a(b-c)}$

$\tt{=a^2+b^2-2ab+a^2+c^2-2ac+b^2-c^2-2ab+2ac}$

$\tt{=a^2+b^2-2ab+a^2+b^2-2ab}$

$\tt{=2\left(a^2+b^2-2ab\right)}$

$\tt{=2\left(a-b\right)^2\ge0}$