Let $a,b,c,d,e,f>0$ satisfying $a+b+c+d+e+f=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f}$ . Prove $$ab+bc+cd+da+ac+bd+ae+be+ce+de+af+bf+cf+df+ef+10\sqrt{abcdef}\geq25$$
Attempt: Prove that if $k>10$ then the inequality $$ab+bc+cd+da+ac+bd+ae+be+ce+de+af+bf+cf+df+ef+k\sqrt{abcdef}\geq15+k$$is not always true.
Also, I tried $$a+b+c+d+e+f\geq\frac{36}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f}}=\frac{36}{a+b+c+f+e+f}.$$ Also, $(a+b+c+d+e+f)^2$ appears $ab+ac+ad+...+ec+ed+ef.$
Let $a+b+c+d+e+f= const$ and $a^2+b^2+c^2+d^2+e^2+f^2=const.$
Thus, by the Vasc's EV Method : https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf (corollary 1.8 (b))
it's enough to prove our inequality for equality case of five variables.
Now, let $b=c=d=e=f$ and $a=xb.$
Thus, the condition gives $$a+5b=\frac{1}{a}+\frac{5}{b}$$ and we need to p-rove that $$5ab+10b^2+10\sqrt{ab^5}\geq25$$ or $$x+2+2\sqrt{\frac{5x+1}{x+5}}\geq\frac{5x(x+5)}{5x+1}$$ or $$\sqrt{\frac{5x+1}{x+5}}\geq\frac{7x-1}{5x+1}$$ and the rest is smooth.