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# Inequality 57 (George Basdekis)

Problem:

If $x_1,x_2,\ldots, x_n$ are non-negative real numbers with sum equal to $1$ then find the maximum value of the expression

$\displaystyle P(x_1,x_2,\ldots,x_n)=x_1^4x_2^2+x_2^4x_3^2+\cdots+x_n^4x_1^2+n^{3(n-1)}x_1^4x_2^4\cdots x_n^4.$

Solution:

First we may, without loss of generality, assume that $x_1=\max\{x_1,x_2,\ldots,x_n\}.$ Now, we shall prove that

$\displaystyle P(x_1,x_2,\ldots,x_n)\leq P(x_1,x_2+x_3+\cdots+x_n,0,0,\ldots,0).$

It holds that

\displaystyle \begin{aligned} P(x_1&,x_2+x_3+\cdots+x_n,0,0,\ldots,0)=x_1^4(x_2+x_3+\cdots+x_n)^2\geq\\&\geq 2x_1^4(x_2x_3+x_3x_4+\cdots+x_nx_1)+x_1^4(x_2^2+x_n^2)\\&\geq (x_1^4x_2x_3+x_1^4x_3x_4+\cdots+x_1^4x_nx_1)+(x_2^4x_3^2+\cdots+x_{n-1}^4x_n^2)+x_1^4x_2^2+x_n^4x_1^2\\&\geq x_1^4x_2x_3+(x_1^4x_2^2+x_1^4x_3^2+x_2^4x_3^2+\cdots+x_n^4x_1^2).\end{aligned}

Therefore, we only need to prove that

$\displaystyle x_1^4x_2x_3\geq n^{3(n-1)}x_1^4x_2^4\cdots x_n^4.$

For $n=3$ we get

$\displaystyle x_1^4x_2x_3\geq 2^6x_1^4x_2^4x_3^4$

or equivalently the above Inequality reduces to the $\displaystyle x_2^3x_3^3\leq\frac{1}{2^6}.$ But this one holds due to the AM-GM Inequality since we have

$\displaystyle x_2^3x_3^3\leq\left[\frac{x_2+x_3}{2}\right]^6\leq\left[\frac{x_1+x_2+x_3}{2}\right]^6=\frac{1}{2^6}.$

Now, for $n>3$ on the other side, we only need to show that

$\displaystyle n^{3(n-1)}x_1^4x_2^4\cdots x_n^4\leq x_1^4x_2x_3$

or

$\displaystyle x_2^3x_3^3x_4^4\cdots x_n^4\leq\frac{1}{n^{3(n-1)}}.$

But, again, from the AM-GM Inequality we acquire

\displaystyle \begin{aligned}x_2^3x_3^3x_4^4\cdots x_n^4\leq (x_2x_3\cdots x_n)^3&\leq\left[\frac{x_2+x_3+\cdots+x_n}{n-1}\right]^{3(n-1)}\\&\leq\left[\frac{x_1+x_2+\cdots+x_n}{n-1}\right]^{3(n-1)}\\&=\frac{1}{n^{3(n-1)}}.\end{aligned}

So, finally, we are left to find the maximum value of the expression

$\displaystyle P(x_1,x_2+x_3+\cdots+x_n,0,0,\ldots,0)=x_1^4(1-x_1)^2.$

We have according to the AM-GM Inequality

\displaystyle \begin{aligned} x_1^4(1-x_1)^2&=4^4\cdot 2^2\cdot\frac{x_1}{4}\cdot\frac{x_1}{4}\cdot\frac{x_1}{4}\cdot\frac{x_1}{4}\cdot\frac{1-x_1}{2}\cdot\frac{1-x_1}{2}\\&\leq 4^4\cdot 2^2\cdot\left[\frac{4\cdot\frac{x_1}{4}+2\cdot\frac{1-x_1}{2}}{4+2}\right]^6\\&=\frac{16}{729}.\end{aligned}

Thus, the maximum value of the expression $P$ is equal to $16/729$ and it is obtained if and only if $x_1=2/3,\, x_2=1/3,\, x_3=x_4=\cdots=x_n=0$ Q.E.D.