# Nice problem due to sir dan sitaru!

• If $a,b,c$ and $d$ are positive real numbers for which $\sqrt3(ad-bc)=bd+ac\neq0$.
Prove that
$d(a+b\sqrt3)-c(b-a\sqrt3)\gt 4 \beta$
where $\beta=\sqrt[4]{abcd}$.

• My solution:
First,note that
$\sqrt3(ad-bc)=ac+bd\Rightarrow d(a\sqrt3-b)=c(a+b\sqrt3)$
that is

(1): $ad\succ bc$ .(2):$a\sqrt3\succ b$ .(3):$\sqrt {abcd}\succ \beta$
So,put this in mind,and write the desired inequality as folllows
$\left(\frac{d^2+c^2}{d}\right)(a+b\sqrt3)\succ 4\beta.$
Now,using (1) and (2) to get
$a\succ\frac{bc}{d}\Rightarrow\left(\frac{d^2+c^2}{d}\right)(a+b\sqrt3)\succ\frac{b}{d}(d^2+c^2)\left(\sqrt3+\frac{c}{d}\right)\Rightarrow\left(\frac{d^2+c^2}{d}\right)(a+b\sqrt3)\succ \frac{b}{d}(d^2+c^2)\left(\frac{b}{a}+\frac{c}{d}\right)$ ..(4)
Now,due to AM-GM inequality we shall obtain that the $Rhs$ of (4) is greater than $4\sqrt{abcd}$,which ends the proof because of (3).And we are done!

Looks like your connection to Expii Forum was lost, please wait while we try to reconnect.