Cfrgtkky

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 $sqrt{ab}+sqrt{bc}
+sqrt{ca}geq 3$

Let $$sumlimits_{cyc}sqrt{24ab+25}<21,$$ $a=kx$, $b=ky$ 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 $$x+y+zgeq xy+xz+yz.$$
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 $$x=sqrt{frac{frac{q^2+5q}{6}cdotfrac{r^2+5r}{6}}{frac{p^2+5p}{6}}}=sqrt{frac{(q^2+5q)(r^2+5r)}{6(p^2+5p)}},$$ we need to prove that
$$sum_{cyc}sqrt{frac{(q^2+5q)(r^2+5r)}{6(p^2+5p)}}geqsum_{cyc}frac{p^2+5p}{6}$$ or
$$sqrt6sum_{cyc}(p^2+5p)(q^2+5q)geqsum_{cyc}(p^2+5p)sqrt{pqrprod_{cyc}(p+5)}.$$
Now, let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.

Thus, we need to prove that
$$sqrt6sum_{cyc}(p^2q^2+5p^2q+5p^2r+25pq)geq$$
$$geqsum_{cyc}(p^2+5)sqrt{w^3(pqr+5(pq+pr+qr)+25(p+q+r)+125)}$$ 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!

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 $f(x)+f(y)+f(z)geq 3f(frac{x+y+z}{3})$

(inequalyti Jensen’s)
But can not prove that $sqrt{ab}+sqrt{bc}
+sqrt{ca}geq 3$

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 $3sqrt{24 +25} = 21$

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

$sqrt{24ab+25} + sqrt{24bc + 25} + sqrt{24ca + 25} ge sqrt{24+25} + sqrt{24+25} + sqrt{24+25} = 21$