note that , the equation given can be rewritten as $2x^2+(a+b+c+d)x+ad+bc=0$ note that $D=(a+b+c+d)^2-8(ad+bc)$ now , let $a=b+p , c=d+q$ with $p,q\ge0$ so , $D=(2b+2d+p+q)^2-8(bd+pd+bd+bq)$ $=(2b+2d+p+q)^2-8(2bd+pd+bq)$ $=4(b+d)^2+(p+q)^2+4(b+d)(p+q)-16bd-8pd-8bq$ $=4(b-d)^2+(p+q)^2+4(b+d)(p+q)-8pd-8bq$ $\ge 4(b-d)^2+(p-q)^2+4(b+d)(p+q)-8pd-8bq$ $=4(d-b)^2+(p-q)^2+4(b+d)(p+q)-8pd-8bq$ $\ge 4(d-b)(p-q)+4(b+d)(p+q)-8pd-8bq$ $=4(dp-dq-bp+bq+bp+bq+dp+dq)-8dp-8bq=0$. hence done!

Isn't it too easy to pose at the IMOTC..I think there is no food for the brain in the problem...

sayantanchakraborty wrote: Isn't it too easy to pose at the IMOTC..I think there is no food for the brain in the problem... It was a fun practice test for juniors two years ago.

Goutham wrote: Let $a\ge b$ and $c\ge d$ be real numbers. Prove that the equation \[(x+a)(x+d)+(x+b)(x+c)=0\]has real roots. Note that \[(a-b)(c-d)\geq 0\implies ac+bd\geq ad+bc\implies (a+b+c+d)^2\geq 4(a+b)(c+d)=4(ad+bc+ac+bd)\geq 8(ad+bc)\]Hence the discriminant of $P(x)=(x+a)(x+b)+(x+c)(x+d)\geq 0$ thus it has real roots

$2(a+b+c+d)^2 \geq 4(a+b)(c+d) \geq 8(ad+bc)$

Consider $f(x)=(x+a)(x+d)+(x+b)(x+c)$. WLOG $b\ge c$, so $a\ge b\ge c\ge d$. We easily obtain: $$f(-d)=(b-d)(c-d)>0$$and $$f(-b)=(a-b)(d-b)<0,$$and the result follows by IVT.