Shhhh! Don't give me the hint

Simply apply Cauchy-Schwarz on

$\biggl( \dfrac{5}{\sqrt{a}} , \dfrac{6}{\sqrt{b}},\dfrac{9}{\sqrt{c}} , \dfrac{12}{\sqrt{d}} \biggr)$ $\text{and}$ $\Bigl( \sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d} \Bigr).$

Then we have $\displaystyle \biggl(\sum \sqrt{a} \dfrac{k}{\sqrt{a}}\biggr)^2 \leq \biggl( \sum \dfrac{k^2}{a} \biggr) \biggl( \sum a\biggr),$ so $\displaystyle (5+6+9+12)^2 \leq (a+b+c+d)\Bigl(\dfrac{25}{a} +\dfrac{36}{b} +\dfrac{81}{c} +\dfrac{12}{d} \Bigr).$

Thus, the answer is $32^2 = \boxed{1024}.$

The equality occurs where $\frac{5}{a}=\frac{6}{b}=\frac{9}{c}=\frac{12}{d}.$

Can Also be done by AM-HM

$\frac {5 * \frac {a}{5}+6 * \frac {b}{6}+9 * \frac {c}{81}+12 * \frac {d}{12}}{5+6+9+12} \geq \frac {5+6+9+12}{5 * \frac {5}{a}+6 * \frac {6}{b}+9 * \frac {9}{c}+12 * \frac {12}{d}}$

Equality occurs when all terms are equal

i used titu's lemma .direct answer

It is also possible to find the minimum via the AM-GM. You can in fact multiply and get 16 terms. 4 of them are constant and the other 12 can be grouped in six pairs of the kind $36 \dfrac{a}{b} + 25 \dfrac{b}{a}$ You then apply the AM >= GM on each of this pair (for istance you get $36 \dfrac{a}{b} + 25 \dfrac{b}{a} \geq 60$ on the pair considered above and repeat same argument for the other five pairs ) and get the upper bound 1024 (that is indeed attained).

So this problem can be solved by C-S, AM-GM, AM-HM. Can anyone find a proof involving QM-AM inequality????

Also I want to note that even with C-S(or Titu's lemma) the terms $a,b,c,d$ must be positive (at least at first glance) in order to consider the square roots $\sqrt{a}, \sqrt{b},\sqrt{c}$ and $\sqrt{d}$ . So if we want to get a proof for $a,b,c,d$ non zero reals we must find a fitting argument (is not enough to just throw in the fact that the C-S inequality works for every real numbers).

If any body saw squares and randomly tried their root as input , Its meeee !!! Got lucky

I have used cauchy- schwarz inequality to solve this question.