An Inequality from Gazeta Matematica, March 2016 II


Several inequalities with solution by Dan Sitaru and Leo Giugiuc have been just published in Gazeta Matematica (March 2016). The simplest of these is discussed on a separate page. So far it gathered more than ten solutions. I'll try to partially reproduce some of that experience.

an inequality from Gazeta Matematica, March 2016, #3

Proof 1

Define matrix $\displaystyle A=\left(\begin{array}\;x&y&z&t\\t&x&y&z\end{array}\right).\;$ We have

$\displaystyle\begin{align} A\cdot A^t &= \left(\begin{array}\;x&y&z&t\\t&x&y&z\end{array}\right)\cdot \left(\begin{array}\;x&t\\y&x\\z&y\\t&z\end{array}\right)\\ &=\left(\begin{array}\;x^2+y^2+z^2+t^2 & xt+xy+zy+tz\\xt+xy+zy+tz & x^2+y^2+z^2+t^2\end{array}\right)\\ &=\left(\begin{array}{c,c}\;1 & (x+z)(y+t)\\(x+z)(y+t) & 1\end{array}\right). \end{align}$

By Cauchy-Binet theorem, $\text{det}(A\cdot A^t)\ge 0.\;$ Therefore, $[(x+z)(y+y)]^2\le 1^2,\;$ or, $(x+z)(y+y)\le 1.$

Proof 2

$(x-y)^2+(y-z)^2+(x-t)^2+(y-t)^2\ge 0\;$ which simplifies to $x^2+y^2+z^2t^2-xy-yz-xt-yt\ge 0.\;$ This is exactly $(x+z)(y+t) \le 1.$

Proof 3

By the AM-QM inequality, $\displaystyle\frac{x+y+z+t}{4}\le\sqrt{\frac{x^2+y^2+z^2+t^2}{4}}=\frac{1}{2}\;$ so that $x+y+z+t\le 2.\;$ Further, by the AM-GM inequality,

$\displaystyle\begin{align} (x+z)(y+t) &\le \left(\sqrt{\frac{x+z+y+t}{2}}\right)^2\\ &\le \left(\sqrt{\frac{2}{2}}\right)^2\\ &=1. \end{align}$

Proof 4

From $(x+z)^2\le 2(x^2+z^2)\;$ and $(y+t)^2\le 2(y^2+t^2)\;$ we obtain a sequence of inequalities:

$\displaystyle\begin{align} (x+z)^2(y+t)^2 &\le 4(x^2+z^2)(y^2+t^2)\\ &\le 4\left(\frac{x^2+z^2+y^2+t^2}{2}\right)^2\\ &= 4\left(\frac{1}{2}\right)^2\\ &= 4\cdot \frac{1}{4}. \end{align}$

and, therefore, $(x+z)(y+t)\le 1.$

Proof 5

Use Lagrange multipliers to prove that


Let $J=(x+z)(y+t)+\lambda (x^2+y^2+z^2+t^2-1).$

Taking $\displaystyle\frac{\partial J}{\partial x}=\frac{\partial J}{\partial y}=\frac{\partial J}{\partial z}=\frac{\partial J}{\partial t}=\frac{\partial J}{\partial\lambda}=0\;$ yields

$y+t+2x\lambda=0,\\ x+z+2y\lambda=0,\\ y+t+2z\lambda=0,\\ x+z+2t\lambda=0,\\ x^2+y^2+z^2+t^2=1.$

It follows that $x=z\;$ and $y=t.\;$ So that $2(x^2+y^2)=1\;$ and we need to find $\max 2x\cdot 2y.\;$ But, since $2xy\le x^2+y^2,\;$ $4xy\le 2(x^2+z^2)=1,$ with the equality only when $x=y.$

To sum up, $\displaystyle\max_{x^2+y^2+z^2+t^2=1}(x+z)(y+t)=1,$ with the maximum achieved for $\displaystyle x=y=z=t=\frac{1}{2}.$

Proof 6

Sk Rejuan used, as before, the Cauchy-Schwarz inequality:

$\begin{align} [(x+z)(y+t)]^2 &= [xy+yz+zt+tx]^2\\ &\le(x^2+y^2+z^2+t^2)(y^2+z^2+t^2+x^2)\\ &=1. \end{align}$

Proof 7

Daniel Liu put the AM-QM inequality into the fast lane:

$\displaystyle (x+z)(y+t)\le\frac{(x+y+z+t)^2}{4}\le x^2+y^2+z^2+t^2=1.$


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