# $\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ 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 $$\sqrt{ab}+\sqrt{bc} +\sqrt{ca}\geq 3$$

• Consider $a=2,b=2, c\to 0^+ \implies \sqrt{ab} + \sqrt{bc}+\sqrt{ca} \to 2$... – Macavity Feb 3 at 8:59
• @Macavity, if $a=2,b=2$, then $c=0\not >0$ from the given constraint. Better if $a=b=1.99$ and $c\approx 0.007$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\approx 2.22<3$. – farruhota Feb 3 at 9:16

## 3 Answers

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 $$0 But the condition gives $$x+y+z=k(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 $$0 and after squaring of the both sides we need to prove that $$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!

• too great . You are very well . Happy a day – Hai Smit Feb 3 at 10:25

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 $$3\sqrt{24 +25} = 21$$

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

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

• $a,b,c>0$ and can not integers – Hai Smit Feb 3 at 8:57
• Ok. By 'can not' I guess you mean that non-integer values are allowed. – mrblewog Feb 3 at 9:00
• Example $a=2,c=1, b=\frac{1}{2}$ then $LHS=21,626>21$ – Hai Smit Feb 3 at 9:04
• Yup, your example is fine. – mrblewog Feb 3 at 9:05
• So has @farruhota provided you with a counterexample when some of $a,b,c$ are non-integer? – mrblewog Feb 3 at 9:18