Prove that $2\sum\limits_{cyc}{} \frac{1}{a}+9k^2\sum\limits_{cyc}{} \frac{1}{a+b}+ t^4\sum\limits_{cyc }^{}\frac{1}{a^2+b^2}$

Question

prove that :if $a,b,c,t,k,u>0$ and $a^2+b^2+c^2=3u^2$ then :

$2\sum_{cyc}{} \frac{1}{a}+9k^2\sum_{cyc}{} \frac{1}{a+b}+ t^4\sum_{cyc }^{}\frac{1}{a^2+b^2}\geq \frac{3(2+3k+t^2)^2}{2u(u+2)}.$

my attempt:

we know this:$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\leq \sqrt{\frac{a^2+b^2+c^2}{3}}=u$

so $2\sum_{cyc}{} \frac{1}{a}\geq \frac{6}{u}= \frac{3.4}{2u}$

and after using CBC inequality we will get :

$9k^2\sum_{cyc}{} \frac{1}{a+b}\geq \frac{81k^2}{2(a+b+c)}\geq\frac{81k^2}{2\sqrt{(a^2}+b^2+c^2)3}=\frac{81k^2}{2.3u} $

and $ t^4\sum_{cyc }^{}\frac{1}{a^2+b^2}\geq \frac{9t^4}{2(a^2+b^2+c^2)}=\frac{3t^4}{2.u^2}$

so $2\sum_{cyc}{} \frac{1}{a}+9k^2\sum_{cyc}{} \frac{1}{a+b}+ t^4\sum_{cyc }^{}\frac{1}{a^2+b^2}\geq \frac{3.4}{2u} +\frac{81k^2}{2.3u} +\frac{3t^4}{2.u^2}=3(\frac{4}{2u} +\frac{9k^2}{2u} +\frac{t^4}{2.u^2})\geq \frac{3(2+3k+t^2)^2}{4u+2.u^2}=\frac{3(2+3k+t^2)^2}{2u(u+2)}.$

so finally:

$2\sum_{cyc}{} \frac{1}{a}+9k^2\sum_{cyc}{} \frac{1}{a+b}+ t^4\sum_{cyc }^{}\frac{1}{a^2+b^2}\geq \frac{3(2+3k+t^2)^2}{2u(u+2)}.$

does my attempt is true?