# An Inequality with Constraint VIII

### Problem

Leo Giugiuc has kindly posted a problem due to Michael Rozenberg (Israel) at the CutTheKnotMath facebook page along with a solution (Solution 1) by Leo Giugiuc and Dan Sitaru:

Let $a,b,c\ge 0\;$ and $ab+bc+ca=a+b+c\gt 0.\;$ Prove that

$\sqrt{24a+1}+\sqrt{24b+1}+\sqrt{24c+1}\ge 15.$

### Lemma

Both solutions exploit the following statement:

Let $a,b,c\ge 0\;$ and $r=ab+bc+ca=a+b+c\gt 0.\;$ Then $r\ge 3.$

Indeed, from $(a-b)^2+(b-c)^2+(c-a)^2\ge 0,\;$ we obtain $ab+bc+ca\le a^2+b^2+c^2\;$ (which also stems from the rearrangement inequality). This is equivalent to $\displaystyle ab+bc+ca\le \frac{(a+b+c)^2}{3}.\;$ Under the conditions of the lemma, this is the same as $\displaystyle a+b+c\le \frac{(a+b+c)^2}{3}\;$ and the assertion follows.

### Solution 1

Let $a=\min\{a,b,c\},\;$ then $bc=\max\{ab,bc,ca\},\;$ such that $bc\ge 1.$ We denote $\sqrt{bc}=p\;$ and $b+c=2s.$ We know that $s\ge p.\;$ We shall consider two case:

**Case 1: $1\le p\le 2$**

Then the statement is equivalent to showing that $f_{p}(s)\ge 15,\;$ where

$\displaystyle f_{p}(s)=\sqrt{24\cdot\frac{2s-p^2}{2s-1}+1}+\sqrt{48s+2+2\sqrt{576p^2+48s+1}}.$

Since $p^2\ge 1,\;$ $f_p\;$ is strictly increasing for $s\ge p\;$ so that $f_{p}(s)\ge f_{p}(p),\;$ implying that we may need just to show than $f_{p}(p)\ge 15.$ I.e., that

$\displaystyle\sqrt{24\cdot\frac{2p-p^2}{2p-1}+1}+\sqrt{48p+2+2\sqrt{576p^2+48p+1}}\ge 15.$

But the left-hand side equals $\displaystyle\sqrt{24\cdot\frac{2p-p^2}{2p-1}+1}+2\sqrt{24p+1}.$ Denote $\sqrt{24p+1}=x,\;$ then $x\in [5,7]\;$ and we need to show that $\displaystyle\frac{-x^4+52x^2-75}{2x^2-26}\ge (15-2x)^2,\;$ which is obvious.

**Case 2: $2\le p$**

In this case,

$\begin{align} \sqrt{24a+1}+\sqrt{24b+1}+\sqrt{24c+1}&\ge 1+\sqrt{48s+2+2\sqrt{576p^2+48s+1}}\\ &\ge 1+2\sqrt{24p+1}\\ &\ge 15. \end{align}$

### Solution 2

We shall consider three cases: $abc\ge 1,\;$ $0\lt abc\lt 1,\;$ and $abc=0.$

**Case 1: $abc\ge 1$**

By the AM-GM inequality

$\sqrt{24a+1}+\sqrt{24b+1}+\sqrt{24c+1}\ge 3\sqrt[6]{(24a+1)(24b+1)(24c+1)}.$

Thus, suffice it to prove $(24a+1)(24b+1)(24c+1)\ge 5^6.$ However, in this case,

$\displaystyle\begin{align} (24a+1)(24b+1)(24c+1)&=24^3abc+24^2\sum_{cyc}ab+24\sum_{cyc}a+1\\ &= 24^3abc+24^2r+24r+1\\ &\ge 24^3+24^2\cdot 3+24\cdot 3+1\\ &=(24+1)^3=25^3=5^6, \end{align}$

as required.

**Case 2: $0\lt abc\lt 1$**

Leo found an error in this part. This is being fixed.

**Case 3: $abc=0$**

Given the condition $ab+bc+ca=a+b+c,\;$ if two of $a,b,c\;$ vanish, so does the third and this would contradict another condition $a+b+c\gt 0.\;$ Therefore, WLOG, we may assume that $a=0\;$ and $b\ne 0 \ne c.\;$ The constraint reduces to $bc=b+c.$ As in the lemma, we obtain $bc=b+c\ge 4.\;$

We have to prove that $1+\sqrt{24b+1}+\sqrt{24c+1}\ge 15,\;$ or $\sqrt{24b+1}+\sqrt{24c+1}\ge 14.\;$ By analogy with the first case, this will follow from $2\left((24b+1)(24c+1\right)^{\frac{1}{4}}\ge 14.\;$ The latter is equivalent to $(24b+1)(24c+1)\ge 7^4.$ But

$\begin{align} (24b+1)(24c+1) &= 24^2bc+24(b+c)+1\\ &\ge 24^2\cdot 4+24\cdot 4+1\\ &=(24\cdot 2+1)^2=49^2=7^4, \end{align}$

as required.

### Remark

It is clear that the problem is a representative of a sequence of problems:

Let $a,b,c\ge 0\;$ and $ab+bc+ca=a+b+c\gt 0,\;$ $n\;$ a positive integer. Prove that

$\sqrt{(n^2-1)a+1}+\sqrt{(n^2-1)b+1}+\sqrt{(n^2-1)c+1}\ge 3n.$

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