$\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$?

Let $$\sum\limits_{cyc}\sqrt{24ab+25}<21,$$ $a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $$\sum_{cyc}\sqrt{24xy+25}=21.$$ Thus, $$\sum_{cyc}\sqrt{24k^2xy+25}<21=\sum_{cyc}\sqrt{24xy+25}$$ or $$\sum_{cyc}\frac{(k^2-1)xy}{\sqrt{24k^2xy+25}+\sqrt{24xy+25}}<0$$ or $$(k^2-1)\sum_{cyc}\frac{xy}{\sqrt{24k^2xy+25}+\sqrt{24xy+25}}<0,$$ 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 $$x+y+z\geq xy+xz+yz.$$ Indeed, let $yz=\frac{p^2+5p}{6},$ $xz=\frac{q^2+5q}{6}$ and $xy=\frac{r^2+5r}{6},$ where $p$, $q$ and $r$ are positives.

Thus, $$\sum_{cyc}\sqrt{4(p^2+5p)+25}=21$$ or $$p+q+r=3$$ and since $$x=\sqrt{\frac{\frac{q^2+5q}{6}\cdot\frac{r^2+5r}{6}}{\frac{p^2+5p}{6}}}=\sqrt{\frac{(q^2+5q)(r^2+5r)}{6(p^2+5p)}},$$ we need to prove that $$\sum_{cyc}\sqrt{\frac{(q^2+5q)(r^2+5r)}{6(p^2+5p)}}\geq\sum_{cyc}\frac{p^2+5p}{6}$$ or $$\sqrt6\sum_{cyc}(p^2+5p)(q^2+5q)\geq\sum_{cyc}(p^2+5p)\sqrt{pqr\prod_{cyc}(p+5)}.$$ Now, let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.

Thus, we need to prove that $$\sqrt6\sum_{cyc}(p^2q^2+5p^2q+5p^2r+25pq)\geq$$ $$\geq\sum_{cyc}(p^2+5)\sqrt{w^3(pqr+5(pq+pr+qr)+25(p+q+r)+125)}$$ or $$\sqrt6(9v^4-6uw^3+5(9uv^2-3w^3)u+75u^2v^2)\geq$$ $$\geq(9u^2-6v^2+15u^2)\sqrt{w^3(w^3+15uv^2+75u^3+125u^3)}$$ or $f(w^3)\geq0,$ where $$f(w^3)=\sqrt3(40u^2v^2+3v^4-7uw^3)u-\sqrt2(4u^2-v^2)\sqrt{w^3(200u^3+15uv^2+w^3)}.$$ 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 $0<p<\frac{3}{2}$ 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!