Dorin Marghidanu's Inequality with Generalization

Source

Dorin  Marghidanu's Inequaity with Generalization, source

Problem

Let $a,b\in\mathbb{R},\,$ $a\lt b,\,$ and $x,y,z\in [a,b].\,$ Prove that

$\displaystyle (x+y)^2+(y+z)^2+(z+x)^2+12ab\le 4(a+b)(x+y+z).$

Where does equality hold? Generalize!

Solution 1

For $u,v\in [a,b],\,$ $(u+v-2a)(u+v-2b)\le 0,\,$ so that

(1)

$(u+v)^2+4ab\le 2(a+b)(u+v)$

Equality occurs when $u=v=a\,$ or $u=v=b.$

If $x_k\in [a,b],\,$ for $k=1,\ldots,n,\,$ then, assuming $x_{n+1}=x_1,\,$ (1) yields

$(x_k+x_{k+1})^2+4ab\le 2(a+b)(x_k+x_{k+1})$

for $k=1,\ldots,n.\,$ Summing up, we get

$\displaystyle \begin{align} \sum_{k=1}^n(x_k+x_{k+1})^2+4nab&\le 2(a+b)\sum_{k=1}^n(x_k+x_{k+1})\\ &=2(a+b)\sum_{k=1}^nx_k. \end{align}$

Equality occurs when all $x_k$ are equal to either $a\,$ or $b.\,$ This solves the problem for $n=3\,$ and provides the required generalization.

Solution 2

First of all note that

(1)

$(x_{i}-a)(x_{i}-b)\leq 0,$

that is

$x_{i}^{2}+ab\leq x_{i}(a+b),$

for every $x_{i}\in \lbrack a,b].$ Adding up for $i=1,2,\ldots n$ we get

(2)

$\displaystyle \sum\limits_{i=1}^{n}x_{i}^{2}+nab\leq (a+b)\sum\limits_{i=1}^{n}x_{i}.$

On the other hand, using the obvious inequality (here and in the following set $x_{n+1}:=x_{1}$)

(3)

$\displaystyle \frac{(x_{i}+x_{i+1})^{2}}{2}\leq x_{i}^{2}+x_{i+1}^{2}$

and adding up for $i=1,2,\ldots n$, we obtain

$\displaystyle \frac{1}{2}\sum\limits_{i=1}^{n}(x_{i}+x_{i+1})^{2}\leq 2\sum\limits_{i=1}^{n}x_{i}^{2}$

that is

(4)

$\displaystyle \frac{1}{4}\sum\limits_{i=1}^{n}(x_{i}+x_{i+1})^{2}\leq \sum\limits_{i=1}^{n}x_{i}^{2},$

so that, using (2) and (4), we conclude that

$\displaystyle \sum\limits_{i=1}^{n}(x_{i}+x_{i+1})^{2}+4nab\leq 4(a+b)\sum\limits_{i=1}^{n}x_{i}.$

For the equality case note that in (3) all the $x_{i}$ must have the same value and from (1) it's clear that this value can only be $a$ or $b.$

Solution 3

Since $x,y,z\in [a,b],\,$ there are $\lambda_1,\lambda_2,\lambda_3\in [0,1]\,$ such that

$x=a\lambda_1+b(1-\lambda_1),\,y=a\lambda_2+b(1-\lambda_2),\,z=a\lambda_3+b(1-\lambda_3).$

Thus

$\displaystyle s=x+y+z=a\sum_{k=1}^3\lambda_k+b(3-\sum_{k=1}^3\lambda_k)=3\left(\lambda a+ b(1-\lambda)\right),$

where $\displaystyle \lambda=\frac{\lambda_1+\lambda_2+\lambda_3}{3}\in [0,1].\,$

$\displaystyle \begin{align} \sum_{cycl}(x+y)^2 &= \sum_{cycl}(s-x)^2=3s^2-2s\cdot s+\sum_{cycl}x^2\\ &=s^2+\sum_{cycl}x^2. \end{align}$

Then $\displaystyle s^2+\sum_{cycl}x^2+12ab\le 4(a+b)\sum_{cycl}x\,$ is equivalent to, say,

$\displaystyle\begin{align} T&=9(\lambda a+(1-\lambda)b)^2+12ab-4(a+b)3(\lambda a+(1-\lambda)b)\\ &\qquad\qquad\qquad\qquad\qquad+\sum_{k=1}^3(\lambda_k a+(1-\lambda_k)b)^2\le 0. \end{align}$

Using Jensen's inequality,

$\displaystyle \begin{align} (\lambda a+(1-\lambda)b)^2&\le \lambda a^2+(1-\lambda)b^2\\ \sum_{k=1}^3(\lambda_k a+(1-\lambda_k)b)^2&\le\left(\sum_{k=1}^3\lambda_k\right) a^2+\left(\sum+{k=1}^n(1-\lambda_k)\right)b^2\\ &=3(\lambda a^2+(1-\lambda)b^2). \end{align}$

Bringing it all together,

$\displaystyle \begin{align} T\,&\le 9(\lambda a^2+(1-\lambda)b^2)+12ab+3(\lambda a^2+(1-\lambda)b^2)\\ &\qquad\qquad\qquad -12(a+b)((\lambda a+(1-\lambda)b)\\ &=12[\lambda a^2+(1-\lambda)b^2-(a+b)(\lambda a+(1-\lambda)b)]+12ab\\ &=12[\lambda(-ab)+(1-\lambda)(-ab)]+12ab=0. \end{align}$

Using the same approach we obtain a generalization:

For $x_k\in[a,b],\,$ $k=1,2,\ldots,n,$

$\displaystyle \sum_{i=1}^n\left(\sum_{k=1}^nx_k-x_i\right)^2+n(n-1)^2ab\le (n-1)^2(a+b)\sum_{k=1}^nx_k.$

Solution 4

$F(x,y,z)=LHS-RHS.\,$ $F\,$ is convex in $x,y,z:\;F"=6.\,$ Hence max is at the endpoints. By symmetry, must be in $\{a,a,a\}\,$ or $\{b,b,b\}.\,$ $F(a,a,a)=F(b,b,b)=0,\,$ QED.

Same argument generalizes to n variables, with $12\,$ replaced with $4n.$

Solution 5

We have

$\displaystyle f=\frac{(x+y)^2+(x+z)^2+(y+z)^2-12 a b}{x+y+z}-4 (a+b)$

$f$ finds its extrema at $\nabla f=0$

$\displaystyle \left( \begin{array}{c} \displaystyle \frac{2 \left(-6 a b+x^2+2 x (y+z)+y z\right)}{(x+y+z)^2} =0\\ \displaystyle \frac{2 (-6 a b+x (2 y+z)+y (y+2 z))}{(x+y+z)^2} =0\\ \displaystyle 2-\frac{2 \left(6 a b+x^2+x y+y^2\right)}{(x+y+z)^2} =0\\ \end{array} \right)$

For solution we get $f^*=\left\{f:x=\sqrt{a} \sqrt{b},y=\sqrt{a} \sqrt{b}, z= \sqrt{a} \sqrt{b}\right\}$, at the geometric average between $a$ and $b$, which is the minimum. $f^*=-4 \left(\sqrt{a} - \sqrt{b}\right)^2$.

We can show that the function $f$ increases away from $f^*$, with maximum $f=0$ in the range where $\left\{x=a,y=a,z=a\right\}$ or $\left\{x=b,y=b,z=b\right\}$.

Acknowledgment

Dorin Marghidanu has kindly posted this problem of his at the CutTheKnotMath facebook page and then commented with his solution (Solution 1) and the links to the solutions by Giulio Franco (Solution 2) and Marian Dinca (Solution 3), the latter offered a different kind of generalization than the other two. Solution 4 is by Lorenzo Villa; Solution 5 is by N. N. Taleb.

|Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62030359

Search by google: