<s>[INST] $\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$? For $a,b,c>0 $ and $a+b+c=ab+bc+ca$ . Prove or disprove that : $\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$
I checked in very many cases. Example :$c=1,
a=2,b=\frac{1}{2}...$ then it’s true, but cannot prove that 
My attempts:
I consider function $ f(x)=\sqrt{24x^2+25}$
And $f’(x)=\frac{24x}{\sqrt{24x^2+25}}$, 
$f’’(x)=\frac{600}{(24x^2+25)(\sqrt{24x^2+25}}>0$.
So $f(x)+f(y)+f(z)\geq 3f(\frac{x+y+z}{3})$.
But cannot prove that $\sqrt{ab}+\sqrt{bc}
+\sqrt{ca}\geq 3$ 
 [/INST] What are the allowed values of $a,b,c$? If they are integers greater than zero then how about this:
Firstly $a = b = c = 1$ meets your constraint, and with these values $3\sqrt{24 +25} = 21$ 
By my assumption that they are positive integers any valid $a,b$ would satisfy $\sqrt{24ab + 25} \ge \sqrt{24 + 25} = 7$
And since the problem is symmetric in $a,b,c$, we can say that 
$\sqrt{24ab+25} + \sqrt{24bc + 25} + \sqrt{24ca + 25} \ge \sqrt{24+25} + \sqrt{24+25} + \sqrt{24+25} = 21$
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