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For $a,b,c>0 $ and $a+b+c=ab+bc+ca$ . Prove or disprove that : $sqrt24ab+25+sqrt24bc+25+sqrt24ca+25geq 21$

I checked in very many cases. Example :$c=1,
a=2,b=frac12...$ then it’s true, but cannot prove that

My attempts:

I consider function $ f(x)=sqrt24x^2+25$

And $f’(x)=frac24xsqrt24x^2+25$,

$f’’(x)=frac600(24x^2+25)(sqrt24x^2+25>0$.

So $f(x)+f(y)+f(z)geq 3f(fracx+y+z3)$.

But cannot prove that $sqrtab+sqrtbc
+sqrtcageq 3$

Let $$sumlimits_cycsqrt24ab+25<21,$$ $a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $$sum_cycsqrt24xy+25=21.$$
Thus,
$$sum_cycsqrt24k^2xy+25<21=sum_cycsqrt24xy+25$$ or
$$sum_cycfrac(k^2-1)xysqrt24k^2xy+25+sqrt24xy+25<0$$ or
$$(k^2-1)sum_cycfracxysqrt24k^2xy+25+sqrt24xy+25<0,$$ which gives $$0<k<1.$$
But the condition gives $$x+y+z=k(xy+xz+yz)<xy+xz+yz,$$
which is a contradiction because we'll prove now that $$x+y+zgeq xy+xz+yz.$$
Indeed, let $yz=fracp^2+5p6,$ $xz=fracq^2+5q6$ and $xy=fracr^2+5r6,$ where $p$, $q$ and $r$ are positives.

Thus, $$sum_cycsqrt4(p^2+5p)+25=21$$ or
$$p+q+r=3$$ and since $$x=sqrtfracfracq^2+5q6cdotfracr^2+5r6fracp^2+5p6=sqrtfrac(q^2+5q)(r^2+5r)6(p^2+5p),$$ we need to prove that
$$sum_cycsqrtfrac(q^2+5q)(r^2+5r)6(p^2+5p)geqsum_cycfracp^2+5p6$$ or
$$sqrt6sum_cyc(p^2+5p)(q^2+5q)geqsum_cyc(p^2+5p)sqrtpqrprod_cyc(p+5).$$
Now, let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.

Thus, we need to prove that
$$sqrt6sum_cyc(p^2q^2+5p^2q+5p^2r+25pq)geq$$
$$geqsum_cyc(p^2+5)sqrtw^3(pqr+5(pq+pr+qr)+25(p+q+r)+125)$$ or
$$sqrt6(9v^4-6uw^3+5(9uv^2-3w^3)u+75u^2v^2)geq$$
$$geq(9u^2-6v^2+15u^2)sqrtw^3(w^3+15uv^2+75u^3+125u^3)$$ or $f(w^3)geq0,$ where
$$f(w^3)=sqrt3(40u^2v^2+3v^4-7uw^3)u-sqrt2(4u^2-v^2)sqrtw^3(200u^3+15uv^2+w^3).$$
We see that $f$ decreases, which says that it's enough to prove the last inequality for the maximal value of $w^3$, which happens for equality case of two variables.

Now, let $q=p$ and $r=3-2p$, where $0<p<frac32$ and after squaring of the both sides we need to prove that
$$p^3(p-1)^2(p+5)^2(224-165p+60p^2-8p^3)geq0,$$ which is true because
$$224-165p+60p^2-8p^3=224-165p+48p^2+4p^2(3-2p)geq224-165p+48p^2>0.$$
Done!

I consider function $ f(x)=sqrt24x^2+25$
And $f’(x)=frac24xsqrt24x^2+25$,
$f’’(x)=frac600(24x^2+25)sqrt24x^2+25>0$
So $f(x)+f(y)+f(z)geq 3f(fracx+y+z3)$

(inequalyti Jensen’s)
But can not prove that $sqrtab+sqrtbc
+sqrtcageq 3$

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 $3sqrt24 +25 = 21$

By my assumption that they are positive integers any valid $a,b$ would satisfy $sqrt24ab + 25 ge sqrt24 + 25 = 7$

And since the problem is symmetric in $a,b,c$, we can say that

$sqrt24ab+25 + sqrt24bc + 25 + sqrt24ca + 25 ge sqrt24+25 + sqrt24+25 + sqrt24+25 = 21$