Birth of an Inequality

Problem

Dan Sitaru has kindly posted a problem at the CutTheKnotMath facebook page, with two solutions - Solution 1 and Solution 2 - both by Leo Giugiuc and Dan Sitaru. The solutions are very much related, both follow the same process, the first backwards, the second forward. It is interesting how much more we learn from the second solution.

Prove that, for $a,b,c\in\mathbb{R}$

$\displaystyle 3(a^2+b^2+c^2)^2\ge 24abc\sqrt[3]{abc}+\sum_{cyc}(a^2+b^2-c^2)^2.$

Solution 1

Expanding the two expressions in parentheses we reduce the inequality to

$8(a^2b^2+b^2c^2+c^2a^2)\ge 24(abc)^{\frac{4}{3}}$

same as $a^2b^2+b^2c^2+c^2a^2\ge 3\sqrt[3]{a^2b^2c^2}\;$ which is shown in one step by the AM-GM inequality.

Solution 2

Successively,

$(a^2+b^2+c^2)^2=(a^2+b^2-c^2)^2+4b^2c^2+4a^2c^2,\\ (a^2+b^2+c^2)^2=(b^2+c^2-a^2)^2+4a^2b^2+4a^2c^2,\\ (a^2+b^2+c^2)^2=(b^2+c^2-a^2)^2+4a^2b^2+4b^2c^2.$

Adding this up gives

$\displaystyle\begin{align} 3(a^2+b^2+c^2)^2&=\sum_{cyc}(a^2+b^2-c^2)^2+8(a^2b^2+b^2c^2+c^2a^2)\\ &\ge \sum_{cyc}(a^2+b^2-c^2)^2+8\cdot 3\sqrt[3]{a^4b^4c^4}\\ &\ge \sum_{cyc}(a^2+b^2-c^2)^2+24abc\sqrt[3]{abc}.\\ \end{align}$

Extra

The three identities at the beginning of Solution 2 beg for a generalization. First, they remind of a simpler $(a^2+b^2)^2=(a^2-b^2)^2 + 4a^2b^2\;$ which leads to

$\displaystyle 2(a^2+b^2)^2\ge\sum_{cyc}(a^2-b^2)^2+8a^2b^2=\sum_{cyc}(a^2-b^2)^2+8(ab)^{\frac{4}{2}}.$

In general, for an integer $n\ge 2,$ and $j,\;$ $1\le j\le n,$

$\displaystyle \left(\sum_{i=1}^{n}a_{i}^2\right)^2=\left(\sum_{i=1}^{n}(-1)^{[i\ne j]}a_{i}^2\right)^2+4a_{j}^{2}\sum_{i=1,i\ne j}^{n}a_{i}^2,$

where the expression $[i\ne j]$ is defined by

$[i\ne j]=\begin{cases} 1,& \text{if}\;i= j\\ 2,& \text{if}\;i\ne j. \end{cases}$

Summing up we obtain

$\displaystyle\begin{align} n\left(\sum_{i=1}^{n}a_{i}^2\right)^2&=\sum_{j=1}^{n}\left(\sum_{i=1}^{n}(-1)^{[i\ne j]}a_{i}^2\right)^2+4\sum_{j=1}^{n}a_{j}^{2}\sum_{i=1,i\ne j}^{n}a_{i}^2,\\ &=\sum_{j=1}^{n}\left(\sum_{i=1}^{n}(-1)^{[i\ne j]}a_{i}^2\right)^2+8\sum_{i\lt j}^{n}a_{i}^{2}a_{j}^{2}\\ &\ge\sum_{j=1}^{n}\left(\sum_{i=1}^{n}(-1)^{[i\ne j]}a_{i}^2\right)^2+8\cdot\frac{n(n-1)}{2}\cdot\sqrt[\displaystyle\frac{n(n-1)}{2}]{\prod_{i=1}^{n}a_{i}^{2(n-1)}}\\ &=\sum_{j=1}^{n}\left(\sum_{i=1}^{n}(-1)^{[i\ne j]}a_{i}^2\right)^2+4n(n-1)\left(\prod_{i=1}^{n}a_{i}\right)^{\displaystyle\frac{4}{n}}. \end{align}$

For positive, $a_i,\;$ $i=1,\ldots,n,$

$\displaystyle n\left(\sum_{i=1}^{n}a_{i}\right)^2\ge \sum_{j=1}^{n}\left(\sum_{i=1}^{n}(-1)^{[i\ne j]}a_{i}\right)^2+4n(n-1)\left(\prod_{i=1}^{n}a_{i}\right)^{\displaystyle\frac{2}{n}}.$

 

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