# Area Inequality in Three Triangles

### Solution 1

First, recollect that $\displaystyle\sum_{cycl}a^2\ge 4\sqrt{3}S.\,$ By the Cauchy-Schwarz inequality,

$\displaystyle (\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2 \le (S_1+S_2+S_3)(1+1+1).$

From here,

\displaystyle\begin{align} \sum_{cycl}a_1^2+\sum_{cycl}a_2^2+\sum_{cycl}a_3^2 &\ge 4\sqrt{3}(S_1+S_2+S_3)\\ &\ge 4\sqrt{3}\frac{1}{3}(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2\\ &\gt 2(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2\\ \end{align}

### Solution 2

By the Cauchy-Schwarz inequality, $\displaystyle \left(\sum_{k=1}^3\sqrt{S_k}\right)^2\le 3\sum_{k=1}^3S_k.\,$ Also, $abc\ge 8(s-a)(s-b)(s-c),\,$ where $2s=a+b+c.\,$ It follows that $\sqrt{abc(a+b+c)}\ge 4S.\,$ By the AM-GM inequality, $xy+yx+yz\ge\sqrt{3xyz(x+y+z)}.\,$ Combining everything and using the Rearrangement inequality,

\displaystyle\begin{align} \sum_{k=1}^3S_k &\le \frac{1}{4\sqrt{3}}\left(\sum_{k=1}^3(a_kb_k+b_kc_k+c_ka_k)\right)\\ &\le \frac{1}{4\sqrt{3}}\left(\sum_{k=1}^3(a_k^2+b_k^2+c_k^2)\right), \end{align}

implying

\displaystyle\begin{align} 2(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2 &\le 6\left(\sum_{k=1}^3S_k\right)\\ &\le \frac{\sqrt{3}}{2}\left(\sum_{k=1}^3(a_k^2+b_k^2+c_k^2)\right)\\ &\lt \sum_{cycl}a_1^2+\sum_{cycl}a_2^2+\sum_{cycl}a_3^2. \end{align}

### Solution 3

By the Cauchy-Schwarz inequality, $2(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2\le 6(S_1+S_2+S_3).\,$ We shall prove that in any $\Delta ABC,$

$\displaystyle\sum_{cycl}ab\ge 4\sqrt{3}S.$

The latter is equivalent to $s^2+r(4R+r)\ge 4\sqrt{3}S.\,$ But

$\displaystyle s^2+r(4R+r)\ge s(3\sqrt{3}r)+r(s\sqrt{3}),$

because $s\ge 3\sqrt{3}r\,$ and $4R+r\ge s\sqrt{3}.\,$ Thus, $s^2+r(4R+r)\ge 4\sqrt{3}rs=4\sqrt{3}S,\,$ with a conclusion that

\displaystyle\begin{align} 6(S_1+S_2+S_3) &\le 4\sqrt{3}(S_1+S_2+S_3)\\ &\le\sum_{cycl}(a_1b_1+a_2b_2+a_3b_3)\\ &\le\sum_{cycl}(a_1^2+a_2^2+a_3^2). \end{align}

Please note that we proved a stronger inequality than required: $\displaystyle 2(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2\lt\sum_{cycl}(a_1b_1+a_2b_2+a_3b_3).$

### Solution 4

$S=sr\,$ and $\displaystyle\left(\frac{s}{3\sqrt{3}}\right)\ge r,\,$ i.e., $\displaystyle S\le\frac{s^2}{3\sqrt{3}}.\,$ With the AM-QM inequality, $\displaystyle S\le \frac{1}{4\sqrt{3}}\sum_{cycl}a^2.$

\displaystyle\begin{align} 2\left(\sum_{k=1}^3\sqrt{S_k}\right)^2 &\le\frac{2}{4\sqrt{3}}\left[\sqrt{\sum_{cycl}a_1^2}+\sqrt{\sum_{cycl}a_2^2}+\sqrt{3}\sum_{cycl}a_3^2\right]\\ &\le\frac{1}{2\sqrt{3}}\left(3\sum_{cycl}a_1^2+3\sum_{cycl}a_2^2+3\sum_{cycl}a_3^2\right)\\ &=\frac{\sqrt{3}}{2}\sum_{cycl}(a_1^2+a_2^2+a_3^2)\\ &\lt\sum_{cycl}(a_1^2+a_2^2+a_3^2). \end{align}

### Remark

The required inequality is rather weak. Each of the available proofs established a stronger inequality:

Prove that in three acute triangles $A_1B_1C_1,\,$ $A_2B_2C_2,\,$ $A_3B_3C_3,\,$

$\displaystyle 4(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2 \le \sqrt{3}\left(\sum_{cycl}a_1^2 + \sum_{cycl}a_2^2 + \sum_{cycl}a_3^2\right).$

Solution 3 provided a further refinement:

Prove that in three acute triangles $A_1B_1C_1,\,$ $A_2B_2C_2,\,$ $A_3B_3C_3,\,$

$\displaystyle 4(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2 \le \sqrt{3}\left(\sum_{cycl}a_1b_1 + \sum_{cycl}a_2b_2 + \sum_{cycl}a_3b_3\right).$

Equality is achieved for three equal equilateral triangles.

### Acknowledgment

The inequality (Romanian Mathematical Magazine) has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. Solution 1 is by Kevin Soto Palacios (Peru); Solution 2 is by Diego Alvariz (India); Solution 3 is by Soumava Chakraborty (India); Solution 4 is by Myagmarsuren Yadamsuren (Mongolia).